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1.
\(4\left(1-cos^23x\right)+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}-4=0\)
\(\Leftrightarrow-4cos^23x+2\left(\sqrt{3}+1\right)cos3x-\sqrt{3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cos3x=-\frac{1}{2}\\cos3x=\frac{\sqrt{3}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\frac{2\pi}{9}+\frac{k2\pi}{3}\\x=\pm\frac{\pi}{18}+\frac{k2\pi}{3}\end{matrix}\right.\)
2.
\(\Leftrightarrow\frac{\sqrt{3}-1}{2\sqrt{2}}sinx-\frac{\sqrt{3}+1}{2\sqrt{2}}cosx=-\frac{\sqrt{3}-1}{2\sqrt{2}}\)
\(\Leftrightarrow sin\left(x-\frac{5\pi}{12}\right)=-cos\left(\frac{5\pi}{12}\right)\)
\(\Leftrightarrow sin\left(x-\frac{5\pi}{12}\right)=sin\left(-\frac{\pi}{12}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{5\pi}{12}=-\frac{\pi}{12}+k2\pi\\x-\frac{5\pi}{12}=\frac{13\pi}{12}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
3.
Nhận thấy \(cosx=0\) ko phải nghiệm, chia 2 vế cho \(cos^2x\)
\(3tan^2x+8tanx+8\sqrt{3}-9=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-\sqrt{3}\\tanx=\frac{3\sqrt{3}-8}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{3}+k2\pi\\x=arctan\left(\frac{3\sqrt{3}-8}{3}\right)+k2\pi\end{matrix}\right.\)
4.
\(\Leftrightarrow sin\left(x-120^0\right)=-cos\left(2x\right)=sin\left(2x-90^0\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-90^0=x-120^0+k360^0\\2x-90^0=300^0-x+k360^0\end{matrix}\right.\)
\(\Leftrightarrow...\)
5.
\(\Leftrightarrow\frac{1}{2}-\frac{1}{2}cos2x=\frac{1}{2}-\frac{1}{2}cos6x\)
\(\Leftrightarrow cos6x=cos2x\)
\(\Leftrightarrow\left[{}\begin{matrix}6x=2x+k2\pi\\6x=-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
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 \(cos^32x\)
\(\frac{8sin^22x}{cos^22x}=\frac{\sqrt{3}sin2x}{cos2x}.\frac{1}{cos^22x}+\frac{1}{cos^22x}\)
\(\Leftrightarrow8tan^22x=\sqrt{3}tan2x\left(1+tan^22x\right)+1+tan^22x\)
\(\Leftrightarrow\sqrt{3}tan^32x-7tan^22x+\sqrt{3}tan2x+1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\frac{1}{\sqrt{3}}\\tanx=\sqrt{3}-2\\tanx=\sqrt{3}+2\end{matrix}\right.\)
\(\Leftrightarrow...\)
a.
\(cos\left(3x-\frac{\pi}{6}\right)=sin\left(2x+\frac{\pi}{3}\right)\)
\(\Leftrightarrow cos\left(3x-\frac{\pi}{6}\right)=cos\left(\frac{\pi}{6}-2x\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\frac{\pi}{6}=\frac{\pi}{6}-2x+k2\pi\\3x-\frac{\pi}{6}=2x-\frac{\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\cos3x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\cos2x\ne\frac{1}{2}\end{matrix}\right.\)
\(tan3x-tanx=0\)
\(\Leftrightarrow\frac{sin3x}{cos3x}-\frac{sinx}{cosx}=0\)
\(\Leftrightarrow sin3x.cosx-cos3x.sinx=0\)
\(\Leftrightarrow sin2x=0\)
\(\Leftrightarrow2sinx.cosx=0\)
\(\Leftrightarrow sinx=0\Leftrightarrow x=k\pi\)
c.
\(\Leftrightarrow\frac{1}{2}+\frac{1}{2}cos\left(2x-\frac{2\pi}{5}\right)=\frac{1}{2}-\frac{1}{2}cos\left(4x+\frac{8\pi}{5}\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=-cos\left(4x+\frac{3\pi}{5}+\pi\right)\)
\(\Leftrightarrow cos\left(2x-\frac{2\pi}{5}\right)=cos\left(4x+\frac{3\pi}{5}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+\frac{3\pi}{5}=2x-\frac{2\pi}{5}+k2\pi\\4x+\frac{3\pi}{5}=\frac{2\pi}{5}-2x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
d.
\(\Leftrightarrow cos^2\left(2x-1\right)=0\)
\(\Leftrightarrow cos\left(2x-1\right)=0\)
\(\Leftrightarrow x=\frac{\pi}{4}+\frac{1}{2}+\frac{k\pi}{2}\)
7.
\(\Leftrightarrow\left[{}\begin{matrix}2x-40^0=60^0+k360^0\\2x-40^0=120^0+n360^0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=50^0+k180^0\\x=80^0+n180^0\end{matrix}\right.\)
Do \(-180^0\le x\le180^0\Rightarrow\left\{{}\begin{matrix}-180^0\le50^0+k180^0\le180^0\\-180^0\le80^0+n180^0\le180^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}-\frac{23}{18}\le k\le\frac{13}{18}\\-\frac{13}{9}\le n\le\frac{5}{9}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}k=\left\{-1;0\right\}\\n=\left\{-1;0\right\}\end{matrix}\right.\)
\(\Rightarrow x=\left\{-130^0;50^0;-100^0;80^0\right\}\)
8.
\(\Leftrightarrow sinx=-\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k2\pi\\x=\frac{5\pi}{4}+k2\pi\end{matrix}\right.\)
5.
\(\Leftrightarrow\frac{\sqrt{2}}{2}sin2x+\frac{\sqrt{2}}{2}cos2x=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin2x.sin\frac{\pi}{4}+cos2x.cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow sin\left(2x+\frac{\pi}{4}\right)=\frac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\frac{\pi}{4}=\frac{\pi}{4}+k2\pi\\2x+\frac{\pi}{4}=\frac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\frac{\pi}{4}+k\pi\end{matrix}\right.\)
6.
\(\Leftrightarrow2sin2x=-1\)
\(\Leftrightarrow sin2x=-\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=-\frac{\pi}{6}+k2\pi\\2x=\frac{7\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{12}+k\pi\\x=\frac{7\pi}{12}+k\pi\end{matrix}\right.\)
a.
\(sinx+cosx+\left(sinx+cosx\right)^2+cos^2x-sin^2x=0\)
\(\Leftrightarrow sinx+cosx+\left(sinx+cosx\right)^2+\left(cosx-sinx\right)\left(sinx+cosx\right)=0\)
\(\Leftrightarrow\left(sinx+cosx\right)\left(1+2cosx\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\\1+2cosx=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\left(x+\frac{\pi}{4}\right)=0\\cosx=-\frac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{\pi}{4}+k\pi\\x=\pm\frac{2\pi}{3}+k2\pi\end{matrix}\right.\)
4.
\(\Leftrightarrow2sinx.cosx-\left(1-2sin^2x\right)+3sinx-cosx-1=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+2sin^2x+3sinx-2=0\)
\(\Leftrightarrow cosx\left(2sinx-1\right)+\left(2sinx-1\right)\left(sinx+2\right)=0\)
\(\Leftrightarrow\left(2sinx-1\right)\left(sinx+cosx+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2sinx-1=0\\sinx+cosx=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx=\frac{1}{2}\\sin\left(x+\frac{\pi}{4}\right)=-\sqrt{2}< -1\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐKXĐ: ...
\(\Leftrightarrow cot\left(\frac{\pi}{4}-x\right)=-\frac{1}{\sqrt{3}}\)
\(\Leftrightarrow\frac{\pi}{4}-x=-\frac{\pi}{3}+k\pi\)
\(\Leftrightarrow x=\frac{7\pi}{12}+k\pi\)
3.
\(\Leftrightarrow cos\frac{x}{4}sinx+sin\frac{x}{4}.cosx-3\left(sin^2x+cos^2x\right)+cosx=0\)
\(\Leftrightarrow sin\left(x+\frac{x}{4}\right)=-cosx\)
\(\Leftrightarrow sin\frac{5x}{4}=sin\left(x-\frac{\pi}{2}\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{5x}{4}=x-\frac{\pi}{2}+k2\pi\\\frac{5x}{4}=\frac{3\pi}{2}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow...\)
pt <=> 1+cos2x + cos3x + cosx = 0
<=> 2cos²x + 2cos2x.cosx = 0
<=> 2cosx.(cos2x + cosx) = 0
<=> 4cosx.cos(3x/2).cos(x/2) = 0 <=>
[cosx = 0
[cos(3x/2) = 0 (tập nghiệm cos3x/2 = 0 chứa tập nghiệm cosx/2 = 0)
<=>
[x = pi/2 + kpi
[3x/2 = pi/2 + kpi
<=>
[x = pi/2 + kpi
[x = pi/3 + 2kpi/3 (k thuộc Z)
sin^2 x + sin^2 2x + sin^2 3x + sin^2 4x =
[1-cos(2x)]/2+ [1-cos(4x)]/2+[1-cos(6x)]/2+[1-cos(8x)]/... =
2- [ cos(2x)+cos(4x)+cos(6x)+cos(8x)]/2 =
2- 1/2· [ cos(2x)+cos(8x)]+cos(4x)+cos(6x)]=
2- 1/2· [ 2·cos(-3x)·cos(5x) + 2· cos(-x)·cos(5x)]=
2- cos(5x)· [cos(3x)+cosx] =
2- cos(5x)· 2·cos(2x)·cosx =
2- 2·cosx·cos(2x)·cos(5x)= 2 <-->
*cosx=0 --> x= pi/2+ k·pi with k thuộc Z or
*cos(2x)=0 --> x= pi/4 + k·pi/2 with k thuộc Z or
* cos(5x)=0 --> x= pi/10+ k·pi/5 with k thuộc Z
2 bài kia chịu cùy :))
cám ơn
2bài cậu giải mình làm theo kiểu gần thế này nhưng sao ra kết quả khác vậy.cả hai luôn