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1.
\(\Leftrightarrow sin^2x\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cos^2x\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(1+cosx\right)\left(sinx+1\right)-2\left(1-cosx\right)=0\)
\(\Leftrightarrow\left(1-cosx\right)\left(sinx+cosx+sinx.cosx-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\Leftrightarrow...\\sinx+cosx+sinx.cosx-1=0\left(1\right)\end{matrix}\right.\)
Xét (1):
Đặt \(sinx+cosx=t\Rightarrow\left[{}\begin{matrix}\left|t\right|\le\sqrt{2}\\sinx.cosx=\frac{t^2-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow t+\frac{t^2-1}{2}-1=0\)
\(\Leftrightarrow t^2+2t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow...\)
2.
\(\Leftrightarrow\sqrt{3}sinx.cosx+\sqrt{2}cos^2x+\sqrt{6}cosx=0\)
\(\Leftrightarrow cosx\left(\sqrt{3}sinx+\sqrt{2}cosx+\sqrt{6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\Leftrightarrow...\\\sqrt{3}sinx+\sqrt{2}cosx=-\sqrt{6}\left(1\right)\end{matrix}\right.\)
Xét (1):
Do \(\sqrt{3}^2+\sqrt{2}^2< \left(-\sqrt{6}\right)^2\) nên (1) vô nghiệm
7.
ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(\frac{\pi}{4}-x\right).sin\left(\frac{\pi}{4}+x\right)\ne0\\cos\left(\frac{\pi}{4}-x\right)cos\left(\frac{\pi}{4}+x\right)\ne0\end{matrix}\right.\)
\(\Leftrightarrow cos2x\ne0\)
Phương trình tương đương:
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{2}-\frac{\pi}{4}-x\right)}=cos^44x\)
\(\Leftrightarrow\frac{sin^42x+cos^42x}{tan\left(\frac{\pi}{4}-x\right).cot\left(\frac{\pi}{4}-x\right)}=cos^24x\)
\(\Leftrightarrow sin^42x+cos^42x=cos^44x\)
\(\Leftrightarrow\left(sin^22x+cos^22x\right)^2-2sin^22x.cos^22x=cos^44x\)
\(\Leftrightarrow1-\frac{1}{2}sin^24x=cos^44x\)
\(\Leftrightarrow2-\left(1-cos^24x\right)=2cos^44x\)
\(\Leftrightarrow2cos^44x-cos^24x-1=0\)
\(\Leftrightarrow\left(cos^24x-1\right)\left(2cos^24x+1\right)=0\)
\(\Leftrightarrow cos^24x-1=0\)
\(\Leftrightarrow sin^24x=0\Leftrightarrow sin4x=0\)
\(\Leftrightarrow2sin2x.cos2x=0\Leftrightarrow sin2x=0\)
\(\Leftrightarrow x=\frac{k\pi}{2}\)
1.
\(cos2x+5=2\left(2-cosx\right)\left(sinx-cosx\right)\)
\(\Leftrightarrow2cos^2x+4=4sinx-4cosx-2sinx.cosx+2cos^2x\)
\(\Leftrightarrow2sinx.cosx-4\left(sinx-cosx\right)+4=0\)
Đặt \(sinx-cosx=t\Rightarrow\left\{{}\begin{matrix}\left|t\right|\le\sqrt{2}\\2sinx.cosx=1-t^2\end{matrix}\right.\)
Pt trở thành:
\(1-t^2-4t+4=0\)
\(\Leftrightarrow t^2+4t-5=0\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-5\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x-\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
\( a){\mathop{\rm sinx}\nolimits} + \cos x = \sqrt 2 \sin 5x\\ \Leftrightarrow \sqrt 2 .\sin \left( {x + \dfrac{\pi }{4}} \right) = \sqrt 2 .\sin 5x\\ \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \sin 5x\\ \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = 5x + k2\pi \\ x + \dfrac{\pi }{4} = \pi - 5x + k2\pi \end{array} \right.\left( {k \in \mathbb {Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l} x = \dfrac{\pi }{{16}} + \dfrac{{k\pi }}{2}\\ x = \dfrac{\pi }{8} + \dfrac{{k\pi }}{3} \end{array} \right.\left( {k \in \mathbb{Z}} \right) \)
\( b)\sqrt 3 \sin 2x + \sin \left( {\dfrac{\pi }{2} + 2x} \right) = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \sin \dfrac{\pi }{2}\cos 2x + \cos \dfrac{\pi }{2}\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + 1.\cos 2x + 0.\sin 2x = 1\\ \Leftrightarrow \sqrt 3 \sin 2x + \cos 2x - 1 = 0\\ \Leftrightarrow 2\sqrt 3 {\mathop{\rm sinxcosx}\nolimits} + 1 - 2{\sin ^2}x - 1 = 0\\ \Leftrightarrow \sqrt 3 {\mathop{\rm sinxcosx}\nolimits} - si{n^2}x = 0\\ \Leftrightarrow {\mathop{\rm sinx}\nolimits} \left( {\sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} } \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} {\mathop{\rm sinx}\nolimits} = 0\\ \sqrt 3 \cos x - {\mathop{\rm sinx}\nolimits} = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \sin \left( {\dfrac{\pi }{3} - x} \right) = 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ \dfrac{\pi }{3} - x = k\pi \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k\pi \\ x = \dfrac{\pi }{3} - k\pi \end{array} \right. \)
Nhiều quá @@ Tách ra đi ><
1d.
Đề ko rõ
1e.
\(\Leftrightarrow\left(4cos^3x-3cosx\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(4cos^2x-3\right)^2.cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left(2cos2x-1\right)^2cos2x-cos^2x=0\)
\(\Leftrightarrow cos^2x\left[\left(2cos2x-1\right)^2.cos2x-1\right]=0\)
\(\Leftrightarrow cos^2x\left(4cos^32x-4cos^22x+cos2x-1\right)=0\)
\(\Leftrightarrow cos^2x\left(cos2x-1\right)\left(4cos^22x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cos2x=1\end{matrix}\right.\) \(\Leftrightarrow...\)
2b.
Đề thiếu
2c.
Nhận thấy \(cos2x=0\) ko phải nghiệm, chia 2 vế cho \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
d/
\(\Leftrightarrow2\left(sinx-cosx\right)\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(sinx-cosx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\left(1\right)\\2\left(1+sinx.cosx\right)=\sqrt{3}cos2x\left(2\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\frac{\pi}{4}=k\pi\Rightarrow x=\frac{\pi}{4}+k\pi\)
\(\left(2\right)\Leftrightarrow2+2sinx.cosx=\sqrt{3}cos2x\)
\(\Leftrightarrow2+sin2x=\sqrt{3}cos2x\)
\(\Leftrightarrow\frac{1}{2}sin2x-\frac{\sqrt{3}}{2}cos2x=-1\)
\(\Leftrightarrow sin\left(2x-\frac{\pi}{3}\right)=-1\)
\(\Leftrightarrow2x-\frac{\pi}{3}=-\frac{\pi}{2}+k2\pi\)
\(\Rightarrow x=-\frac{\pi}{12}+k\pi\)
c/
\(\Leftrightarrow sinx-sin^2x=cosx-cos^2x\)
\(\Leftrightarrow sinx-cosx-\left(sin^2x-cos^2x\right)=0\)
\(\Leftrightarrow sinx-cosx-\left(sinx-cosx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx-cosx\right)\left(1-sinx-cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=0\\1-sinx-cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x-\frac{\pi}{4}\right)=0\\1-\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x-\frac{\pi}{4}\right)=0\\sin\left(x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{\pi}{4}=k\pi\\x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=k2\pi\\x=\frac{\pi}{2}+k2\pi\end{matrix}\right.\)
a) Đk: sinx \(\ne\)0<=>x\(\ne\)k\(\Pi\)
pt<=>\(\sqrt{3}\)(1-cos2x)-cosx=0
<=>\(\sqrt{3}\)[1-(2cos2x-1)]-cosx=0
<=>2\(\sqrt{3}\)-2\(\sqrt{3}\)cos2x-cosx=0
<=>\(\left\{{}\begin{matrix}cosx=\dfrac{\sqrt{3}}{2}\\cosx=-\dfrac{2\sqrt{3}}{3}< -1\left(loai\right)\end{matrix}\right.\)
tới đây bạn tự giải cho quen, chứ chép thì thành ra không hiểu gì thì khổ
b)pt<=>2sin2x+2sin2x=1
<=>2sin2x+2sin2x=sin2x+cos2x
<=>4sinx.cosx+sin2x-cos2x=0
Tới đây là dạng của pt đẳng cấp bậc 2, ta thấy cosx=0 không phải là nghiệm của pt nên ta chia cả hai vế của pt cho cos2x:
pt trở thành:
4tanx+tan2x-1=0
<=>\(\left[{}\begin{matrix}tanx=-2+\sqrt{2}\\tanx=-2-\sqrt{5}\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=arctan\left(-2+\sqrt{5}\right)+k\Pi\\x=arctan\left(-2-\sqrt{5}\right)+k\Pi\end{matrix}\right.\)(k thuộc Z)
Chú ý: arctan tương ứng ''SHIFT tan'' (khi thử nghiệm trong máy tính)
c)Đk: cosx\(\ne\)0<=>x\(\ne\)\(\dfrac{\Pi}{2}\)+kpi
pt<=>cos2x+\(\sqrt{3}\)sin2x=1
<=>1-sin2x+\(\sqrt{3}\)sin2x-1=0
<=>(\(\sqrt{3}\)-1)sin2x=0
<=>sinx=0<=>x=k\(\Pi\)(k thuộc Z)
d)
pt<=>\(\sqrt{3}\)sin7x-cos7x=\(\sqrt{2}\)
Khúc này bạn coi SGK trang 35 người ta giả thích rõ ràng rồi
pt<=>\(\dfrac{\sqrt{3}}{2}\)sin7x-\(\dfrac{1}{2}\)cos7x=\(\dfrac{\sqrt{2}}{2}\)
<=>sin(7x-\(\dfrac{\Pi}{3}\))=\(\dfrac{\sqrt{2}}{2}\)
<=>sin(7x-\(\dfrac{\Pi}{3}\))=sin\(\dfrac{\Pi}{4}\)
Tới đây bạn tự giải nhé, giải ra nghiệm rồi kiểm tra xem nghiệm nào thuộc khoảng ( đề cho) rồi kết luận
Câu d) mình nhầm nhé
<=>sin(7x-\(\dfrac{\Pi}{6}\))=\(\dfrac{\sqrt{2}}{2}\) mới đúng sorry
1.
\(\Leftrightarrow\left[{}\begin{matrix}cos4x=-\frac{\sqrt{3}}{2}\\cos4x=-\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow x=...\)
(Cứ bấm máy giải pt bậc 2 như bt, nó cho 2 nghiệm rất xấu, bạn lưu 2 nghiệm vào 2 biến A; B rồi thoát ra ngoài MODE-1, tính \(\sqrt{A^2}\) và \(\sqrt{B^2}\) sẽ ra dạng căn đẹp của 2 nghiệm, lưu ý dấu so với nghiệm ban đầu)
2.
\(\Leftrightarrow cos4x+1+sin\left(2x-\frac{\pi}{2}\right)=cos2x\)
\(\Leftrightarrow2cos^22x-cos2x=cos2x\)
\(\Leftrightarrow cos^22x-cos2x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\cos2x=1\end{matrix}\right.\)
3.
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left[\frac{\pi}{2}-\left(\frac{\pi}{6}-x\right)\right]=\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}sin\left(x+\frac{\pi}{3}\right)+\frac{\sqrt{3}}{2}cos\left(x+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{3}+\frac{\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow sin\left(x+\frac{2\pi}{3}\right)=\frac{1}{2}\)
\(\Leftrightarrow...\)
4.
\(\Leftrightarrow2cos4x.cos\left(\frac{\pi}{3}\right)+2sin4x.sin\left(\frac{\pi}{3}\right)+4cos2x=-1\)
\(\Leftrightarrow cos4x+\sqrt{3}sin4x+4cos2x+1=0\)
\(\Leftrightarrow2cos^22x+2\sqrt{3}sin2x.cos2x+4cos2x=0\)
\(\Leftrightarrow2cos2x\left(cos2x+\sqrt{3}sin2x+2\right)=0\)
\(\Leftrightarrow cos2x\left(\frac{\sqrt{3}}{2}sin2x+\frac{1}{2}cos2x+1\right)=0\)
\(\Leftrightarrow cos2x\left[sin\left(2x+\frac{\pi}{6}\right)+1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=0\\sin\left(2x+\frac{\pi}{6}\right)=-1\end{matrix}\right.\)
2.1
a.
\(\Leftrightarrow sinx-cosx=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{13\pi}{12}+k2\pi\end{matrix}\right.\)
b.
\(cosx-\sqrt{3}sinx=1\)
\(\Leftrightarrow\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
c.
\(\sqrt{3}sin\dfrac{x}{3}+cos\dfrac{x}{2}=\sqrt{2}\)
Câu này đề đúng không nhỉ? Nhìn thấy có vẻ không đúng lắm
d.
\(cosx-sinx=1\)
\(\Leftrightarrow\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
e.
\(2cosx+2sinx=\sqrt{6}\)
\(\Leftrightarrow2\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=\sqrt{6}\)
\(\Leftrightarrow cos\left(x-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{4}=-\dfrac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{12}+k2\pi\\x=\dfrac{\pi}{12}+k2\pi\end{matrix}\right.\)
f.
\(sin3x+\sqrt{3}cosx=\sqrt{2}\)
Lại 1 câu đề không đúng
g.
\(3sinx-2cosx=2\)
\(\Leftrightarrow\dfrac{3}{\sqrt{13}}sinx-\dfrac{2}{\sqrt{13}}cosx=\dfrac{2}{\sqrt{13}}\)
Đặt \(\dfrac{3}{\sqrt{13}}=cosa\) với \(0< a< \dfrac{\pi}{2}\Rightarrow\dfrac{2}{\sqrt{13}}=sina\)
\(\Rightarrow sinx.cosa-cosx.sina=sina\)
\(\Leftrightarrow sin\left(x-a\right)=sina\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=a+k2\pi\\x-a=\pi-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2a+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
2.2
a.
\(\left(sinx-1\right)\left(1+cosx\right)=cos^2x\)
\(\Leftrightarrow\left(sinx-1\right)\left(1+cosx\right)=1-sin^2x\)
\(\Leftrightarrow\left(sinx-1\right)\left(1+cosx\right)=\left(1-sinx\right)\left(1+sinx\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-1=0\\-1-cosx=1+sinx\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=1\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k2\pi\\sin\left(x+\dfrac{\pi}{4}\right)=-\sqrt{2}< -1\left(vô-nghiệm\right)\end{matrix}\right.\)
b.
\(sin\left(\dfrac{\pi}{2}+2x\right)+\sqrt{3}sin\left(\pi-2x\right)=1\)
\(\Leftrightarrow cos2x+\sqrt{3}sin2x=1\)
\(\Leftrightarrow\dfrac{1}{2}cos2x+\dfrac{\sqrt{3}}{2}sin2x=\dfrac{1}{2}\)
\(\Leftrightarrow cos\left(2x-\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\2x-\dfrac{\pi}{3}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k\pi\\x=k\pi\end{matrix}\right.\)
c.
\(\sqrt{2}\left(cos^4x-sin^4x\right)=cosx+sinx\)
\(\Leftrightarrow\sqrt{2}\left(cos^2x-sin^2x\right)\left(cos^2x+sin^2x\right)=cosx+sinx\)
\(\Leftrightarrow\sqrt{2}\left(cos^2x-sin^2x\right)=cosx+sinx\)
\(\Leftrightarrow\sqrt{2}\left(cosx-sinx\right)\left(cosx+sinx\right)=cosx+sinx\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx+sinx=0\\\sqrt{2}\left(cosx-sinx\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2}cos\left(x-\dfrac{\pi}{4}\right)=0\\2cos\left(x+\dfrac{\pi}{4}\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi\\x+\dfrac{\pi}{4}=\dfrac{\pi}{3}+k2\pi\\x+\dfrac{\pi}{4}=-\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+k2\pi\\x=-\dfrac{7\pi}{12}+k2\pi\end{matrix}\right.\)
d.
\(sin2x+cos2x=\sqrt{2}sin3x\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=\sqrt{2}sin3x\)
\(\Leftrightarrow sin3x=sin\left(2x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=2x+\dfrac{\pi}{4}+k2\pi\\3x=\dfrac{3\pi}{4}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{3\pi}{20}+\dfrac{k2\pi}{5}\end{matrix}\right.\)
e.
\(sinx=\sqrt{2}sin5x-cosx\)
\(\Leftrightarrow sinx+cosx=\sqrt{2}sin5x\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin5x\)
\(\Leftrightarrow sin5x=sin\left(x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=x+\dfrac{\pi}{4}+k2\pi\\5x=\dfrac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{16}+\dfrac{k\pi}{2}\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{3}\end{matrix}\right.\)
f.
\(sin8x-cos6x=\sqrt{3}\left(sin6x+cos8x\right)\)
\(\Leftrightarrow sin8x-\sqrt{3}cos8x=cos6x+\sqrt{3}sin6x\)
\(\Leftrightarrow\dfrac{1}{2}sin8x-\dfrac{\sqrt{3}}{2}cos8x=\dfrac{1}{2}cos6x+\dfrac{\sqrt{3}}{2}sin6x\)
\(\Leftrightarrow sin\left(8x-\dfrac{\pi}{3}\right)=sin\left(6x+\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}8x-\dfrac{\pi}{3}=6x+\dfrac{\pi}{6}+k2\pi\\8x-\dfrac{\pi}{3}=\dfrac{5\pi}{6}-6x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=\dfrac{\pi}{12}+\dfrac{k\pi}{7}\end{matrix}\right.\)
g.
\(cos3x-sinx=\sqrt{3}\left(cosx-sin3x\right)\)
\(\Leftrightarrow cos3x+\sqrt{3}sin3x=\sqrt{3}cosx+sinx\)
\(\Leftrightarrow\dfrac{1}{2}cos3x+\dfrac{\sqrt{3}}{2}sin3x=\dfrac{\sqrt{3}}{2}cosx+\dfrac{1}{2}sinx\)
\(\Leftrightarrow cos\left(3x-\dfrac{\pi}{3}\right)=cos\left(x-\dfrac{\pi}{6}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{\pi}{3}=x-\dfrac{\pi}{6}+k2\pi\\3x-\dfrac{\pi}{3}=\dfrac{\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{2}\end{matrix}\right.\)
h.
\(2sin^2x+\sqrt{3}sin2x=3\)
\(\Leftrightarrow1-cos2x+\sqrt{3}sin2x=3\)
\(\Leftrightarrow\sqrt{3}sin2x-cos2x=2\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sin2x-\dfrac{1}{2}cos2x=1\)
\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{6}\right)=1\)
\(\Leftrightarrow2x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{3}+k\pi\)
i.
\(sin^4x+cos^4\left(x+\dfrac{\pi}{2}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow\left(\dfrac{1-cos2x}{2}\right)^2+\left[\dfrac{1+cos\left(2x+\dfrac{\pi}{2}\right)}{2}\right]^2=\dfrac{1}{4}\)
\(\Leftrightarrow\left(\dfrac{1-cos2x}{2}\right)^2+\left(\dfrac{1-sin2x}{2}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow1-2cos2x+cos^22x+1-2sin2x+sin^22x=1\)
\(\Leftrightarrow3-2\left(sin2x+cos2x\right)=1\)
\(\Leftrightarrow sin2x+cos2x=1\)
\(\Leftrightarrow\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\2x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
2.3
a.
\(\dfrac{\sqrt{3}\left(1-cos2x\right)}{2sinx}=cosx\)
ĐKXĐ: \(x\ne k\pi\)
\(\sqrt{3}\left(1-cos2x\right)=2sinx.cosx\)
\(\Leftrightarrow\sqrt{3}-\sqrt{3}cos2x=sin2x\)
\(\Leftrightarrow sin2x+\sqrt{3}cos2x=\sqrt{3}\)
\(\Leftrightarrow\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}cos2x=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{\sqrt{3}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{3}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\left(loại\right)\\x=\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
b.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(cotx-tanx=\dfrac{cosx-sinx}{sinx.cosx}\)
\(\Leftrightarrow\dfrac{sinx}{cosx}-\dfrac{cosx}{sinx}=\dfrac{cosx-sinx}{sinx.cosx}\)
\(\Leftrightarrow\dfrac{sin^2x-cos^2x}{sinx.cosx}=\dfrac{cosx-sinx}{sinx.cosx}\)
\(\Rightarrow\left(sinx-cosx\right)\left(sinx+cosx\right)=-\left(sinx-cosx\right)\)
\(\Rightarrow\left[{}\begin{matrix}sinx-cosx=0\\sinx+cosx=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=1\\sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k\pi\\x=-\dfrac{\pi}{2}+k2\pi\left(loại\right)\\x=\dfrac{3\pi}{2}+k2\pi\left(loại\right)\end{matrix}\right.\)
c.
ĐKXĐ: \(x\ne\dfrac{k\pi}{2}\)
\(\dfrac{\sqrt{3}}{cosx}+\dfrac{1}{sinx}=4\)
\(\Rightarrow\sqrt{3}sinx+cosx=4sinx.cosx\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx+\dfrac{1}{2}cosx=sin2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{6}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{6}+k2\pi\\2x=\dfrac{5\pi}{6}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
d.
\(\dfrac{1+sinx}{1+cosx}=\dfrac{1}{2}\)
ĐKXĐ: \(x\ne\pi+k2\pi\)
\(2\left(1+sinx\right)=1+cosx\)
\(2sinx+2=1+cosx\)
\(\Leftrightarrow cosx-2sinx=1\)
\(\Leftrightarrow\dfrac{1}{\sqrt{5}}cosx-\dfrac{2}{\sqrt{5}}sinx=\dfrac{1}{\sqrt{5}}\)
Đặt \(\dfrac{1}{\sqrt{5}}=cosa\) với \(0< a< 90^0\)
\(\Rightarrow cosx.cosa-sinx.sina=cosa\)
\(\Leftrightarrow cos\left(x+a\right)=cosa\)
\(\Leftrightarrow\left[{}\begin{matrix}x+a=a+k2\pi\\x+a=-a+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=-2a+k2\pi\end{matrix}\right.\)
e.
\(3cosx+4sinx+\dfrac{6}{3cosx+4sinx+1}=6\)
Đặt \(3cosx+4sinx+1=t\)
\(\Rightarrow t-1+\dfrac{6}{t}=6\)
\(\Rightarrow t^2-7t+6=0\Rightarrow\left[{}\begin{matrix}t=1\\t=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3cosx+4sinx+1=1\\3cosx+4sinx+1=6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3cosx+4sinx=0\\3cosx+4sinx=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{5}cosx+\dfrac{4}{5}sinx=0\\\dfrac{3}{5}cosx+\dfrac{4}{5}sinx=1\end{matrix}\right.\)
Đặt \(\dfrac{3}{5}=cosa\) với \(0< a< \dfrac{\pi}{2}\)
\(\Rightarrow\left[{}\begin{matrix}cosx.cosa+sinx.sina=0\\cosx.cosa+sinx.sina=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}cos\left(x-a\right)=0\\cos\left(x-a\right)=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-a=\dfrac{\pi}{2}+k\pi\\x-a=k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a+\dfrac{\pi}{2}+k2\pi\\x=a+k2\pi\end{matrix}\right.\)
2.4
a.
\(cos7x-\sqrt{3}sin7x=-\sqrt{2}\)
\(\Leftrightarrow\dfrac{1}{2}cos7x-\dfrac{\sqrt{3}}{2}sin7x=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow cos\left(7x+\dfrac{\pi}{3}\right)=-\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}7x+\dfrac{\pi}{3}=\dfrac{3\pi}{4}+k2\pi\\7x+\dfrac{\pi}{3}=-\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{84}+\dfrac{k2\pi}{7}\\x=-\dfrac{13\pi}{84}+\dfrac{k2\pi}{7}\end{matrix}\right.\)
Do \(\dfrac{2\pi}{5}< x< \dfrac{6\pi}{7}\Rightarrow\left[{}\begin{matrix}\dfrac{2\pi}{5}< \dfrac{5\pi}{84}+\dfrac{k2\pi}{7}< \dfrac{6\pi}{7}\\\dfrac{2\pi}{5}< -\dfrac{13\pi}{84}+\dfrac{k2\pi}{7}< \dfrac{6\pi}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}k=2\\k=\left\{2;3\right\}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{53\pi}{84}\\x=\dfrac{5\pi}{12}\\x=\dfrac{59\pi}{84}\end{matrix}\right.\)
2.4
b.
\(4sin^2\dfrac{x}{2}-\sqrt{3}cos2x=1+2cos^2\left(x-\dfrac{3\pi}{4}\right)\)
\(\Leftrightarrow2-2cosx-\sqrt{3}cos2x=1+1+cos\left(2x-\dfrac{3\pi}{2}\right)\)
\(\Leftrightarrow-2cosx-\sqrt{3}cos2x=-sin2x\)
\(\Leftrightarrow\sqrt{3}cos2x-sin2x=-2cosx\)
\(\Leftrightarrow\dfrac{\sqrt{3}}{2}cos2x-\dfrac{1}{2}sin2x=cos\left(\pi-x\right)\)
\(\Leftrightarrow cos\left(2x+\dfrac{\pi}{6}\right)=cos\left(\pi-x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{6}=\pi-x+k2\pi\\2x+\dfrac{\pi}{6}=x-\pi+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5\pi}{18}+\dfrac{k2\pi}{3}\\x=-\dfrac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(0< x< \pi\Rightarrow x=\left\{\dfrac{5\pi}{18};\dfrac{17\pi}{18};\dfrac{5\pi}{6}\right\}\)
2.5
Theo điều kiện có nghiệm của pt lượng giác bậc nhất, các pt có nghiệm khi:
a.
\(m^2+\left(m+1\right)^2\ge m^2\)
\(\Leftrightarrow\left(m+1\right)^2\ge0\) (luôn thỏa mãn)
Vậy pt đã cho có nghiệm với mọi m
b.
\(\left(2m-1\right)^2+\left(m-1\right)^2\ge\left(m-3\right)^2\)
\(\Leftrightarrow4m^2-7\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}m\ge\dfrac{\sqrt{7}}{2}\\m\le-\dfrac{\sqrt{7}}{2}\end{matrix}\right.\)
2.6
a.
\(y=5\left(\dfrac{3}{5}sinx-\dfrac{4}{5}cosx\right)+5\)
Đặt \(\dfrac{3}{5}=cosa\) với \(0< a< \dfrac{\pi}{2}\)
\(\Rightarrow y=5\left(sinx.cosa-cosx.sina\right)+5=5sin\left(x-a\right)+5\)
Do \(-1\le sin\left(x-a\right)\le1\)
\(\Rightarrow0\le y\le10\)
\(y_{min}=0\) khi \(sin\left(x-a\right)=-1\)
\(y_{max}=10\) khi \(sin\left(x-a\right)=1\)
b.
\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)-1\)
Do \(-1\le sin\left(2x+\dfrac{\pi}{4}\right)\le1\)
\(\Rightarrow-\sqrt{2}-1\le y\le\sqrt{2}-1\)
\(y_{max}=\sqrt{2}-1\) khi \(sin\left(2x+\dfrac{\pi}{4}\right)=1\)
\(y_{min}=-\sqrt{2}-1\) khi \(sin\left(2x+\dfrac{\pi}{4}\right)=-1\)
dạ em lộn ạ đề đúng phải là \(\sqrt{3}sin\dfrac{x}{2}+cos\dfrac{x}{2}=\sqrt{2}\) ạ