\(2cos3x\left(2-4sin^2x+1\right)=1\)

\(\Leftrightarrow2cos3x\left(3-4sin^2x\right)=1\)

Nhận thấy \(sinx=0\Leftrightarrow x=k\pi\) không phải nghiệm, nhân 2 vế của pt với \(sinx:\)

\(2cos3x\left(3sinx-4sin^3x\right)=sinx\)

\(\Leftrightarrow2cos3x.sin3x=sinx\)

\(\Leftrightarrow sin6x=sinx\Leftrightarrow\left[{}\begin{matrix}6x=x+k2\pi\\6x=\pi-x+l2\pi\end{matrix}\right.\) (chú ý \(x\ne m.\pi\))

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k2\pi}{5}\\x=\dfrac{\pi}{7}+\dfrac{l2\pi}{7}\end{matrix}\right.\) ; \(x\ne m.\pi\)

Xét trên \(\left[-4\pi;6\pi\right]\): \(\left\{{}\begin{matrix}-4\pi\le\dfrac{k2\pi}{5}\le6\pi\\-4\pi\le\dfrac{\pi+l2\pi}{7}\le6\pi\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}-10\pi\le k\le15\pi\\-13\le l\le20\pi\end{matrix}\right.\)

Vậy tổng các nghiệm:

\(S=\pi\left(\sum\limits^{15}_{k=-10}\dfrac{2k}{5}+\sum\limits^{20}_{l=-13}\dfrac{2l+1}{7}-\sum\limits^6_{m=-4}m\right)=\dfrac{377.\pi}{7}\)

Đáp án C

A

A

\(\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)\)

\(=-\left(cos^2x-\sin^2x\right)=-cos2x\)

Phương trình ban đầu sẽ trở thành: \((2\cos 2x + 5)(-\cos 2x) + 3 = 0\)

=>\(-2\cdot cos^22x-5\cdot cos2x+3=0\)

=>\(-2\cdot cos^22x-6\cdot cos2x+cos2x+3=0\)

=>(cos2x+3)(-2cos2x+1)=0

=>-2cos2x+1=0

=>\(\cos2x=\frac{1}{2}\)

=>\(\left[\begin{array}{l}2x=\frac{\pi}{3}+k2\pi\iff x=\frac{\pi}{6}+k\pi\\ 2x=-\frac{\pi}{3}+k2\pi\iff x=-\frac{\pi}{6}+k\pi\end{array}\right.\quad\)

TH1: \(x=\frac{\pi}{6}+k\pi\)

mà x∈(0;2Ω)

nên \(x\in\left\lbrace\frac{\pi}{6};\frac76\pi\right\rbrace\)

TH2: \(x=-\frac{\pi}{6}+k\pi\)

mà \(x\in\left(0;2\pi\right)\)

nên \(x\in\left\lbrace\frac56\pi;\frac{11}{6}\pi\right\rbrace\)

Tổng các nghiệm là:

\(S=\frac{\pi}{6}+\frac76\pi+\frac56\pi+\frac{11}{6}\pi=4\pi\)

=>2cos2x=pi(loại) hoặc sin x-cosx=0

=>sin x-cosx=0

=>sin(x-pi/4)=0

=>x-pi/4=kpi

=>x=kpi+pi/4

mà x\(\in\left[-pi;pi\right]\)

nên \(x\in\left\{\dfrac{pi}{4};-\dfrac{3}{4}pi\right\}\)

=> D

\(\Leftrightarrow2\left(cos^2x-sin^2x\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)

\(\Leftrightarrow\left(2cosx-2sinx\right)\left(sinx+cosx\right)+sinx.cosx\left(sinx+cosx\right)=m\left(sinx+cosx\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx+cosx=0\left(\text{vô nghiệm trên đoạn xét}\right)\\2cosx-2sinx+sinx.cosx=m\left(1\right)\end{matrix}\right.\) 

Xét (1), đặt \(t=cosx-sinx=\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)\)

\(\Rightarrow\left\{{}\begin{matrix}t\in\left[-1;1\right]\\sinx.cosx=\dfrac{1-t^2}{2}\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow2t+\dfrac{1-t^2}{2}=m\)

Xét hàm \(f\left(t\right)=-\dfrac{1}{2}t^2+2t+\dfrac{1}{2}\) trên \(\left[-1;1\right]\)

\(-\dfrac{b}{2a}=2\notin\left[-1;1\right]\) ; \(f\left(-1\right)=-2\) ; \(f\left(1\right)=2\)

\(\Rightarrow-2\le f\left(t\right)\le2\Rightarrow-2\le m\le2\)

3.

\(\Leftrightarrow\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=cos3x\)

\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=sin\left(\dfrac{\pi}{2}-3x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{2}-3x+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{\pi}{2}+3x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{3}+k\pi\end{matrix}\right.\)

câu 2 mình sửa lại đề bài một chút là: sin(cosx)=1 ạ

1.

\(cos2x-3cosx+2=0\)

\(\Leftrightarrow2cos^2x-3cosx+1=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\pm\dfrac{\pi}{3}+k2\pi\end{matrix}\right.\)

\(x=k2\pi\in\left[\dfrac{\pi}{4};\dfrac{7\pi}{4}\right]\Rightarrow\) không có nghiệm x thuộc đoạn

\(x=\pm\dfrac{\pi}{3}+k2\pi\in\left[\dfrac{\pi}{4};\dfrac{7\pi}{4}\right]\Rightarrow x_1=\dfrac{\pi}{3};x_2=\dfrac{5\pi}{3}\)

\(\Rightarrow P=x_1.x_2=\dfrac{5\pi^2}{9}\)

2.

\(pt\Leftrightarrow\left(cos3x-m+2\right)\left(2cos3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cos3x=\dfrac{1}{2}\left(1\right)\\cos3x=m-2\left(2\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=\pm\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\)

Ta có: \(x=\pm\dfrac{\pi}{9}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=\pm\dfrac{\pi}{9}\)

Yêu cầu bài toán thỏa mãn khi \(\left(2\right)\) có nghiệm duy nhất thuộc \(\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}m-2=0\\m-2=1\\m-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=2\\m=3\\m=1\end{matrix}\right.\)

TH1: \(m=2\)

\(\left(2\right)\Leftrightarrow cos3x=0\Leftrightarrow x=\dfrac{\pi}{6}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=\dfrac{\pi}{6}\left(tm\right)\)

\(\Rightarrow m=2\) thỏa mãn yêu cầu bài toán

TH2: \(m=3\)

\(\left(2\right)\Leftrightarrow cos3x=0\Leftrightarrow x=\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow x=0\left(tm\right)\)

\(\Rightarrow m=3\) thỏa mãn yêu cầu bài toán

TH3: \(m=1\)

\(\left(2\right)\Leftrightarrow cos3x=-1\Leftrightarrow x=\dfrac{\pi}{3}+\dfrac{k2\pi}{3}\in\left(-\dfrac{\pi}{6};\dfrac{\pi}{3}\right)\Rightarrow\left[{}\begin{matrix}x=\pm\dfrac{1}{3}\\x=-1\\x=-\dfrac{5}{3}\end{matrix}\right.\)

\(\Rightarrow m=2\) không thỏa mãn yêu cầu bài toán

Vậy \(m=2;m=3\)

Ta có

\(\begin{array}{l}\sin \left( {x + \frac{\pi }{4}} \right) = \frac{{\sqrt 2 }}{2}\\ \Leftrightarrow \sin \left( {x + \frac{\pi }{4}} \right) = \sin \left( {\frac{\pi }{4}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x + \frac{\pi }{4}{\rm{ }} = {\rm{ }}\frac{\pi }{4} + k2\pi ;k \in Z\\x + \frac{\pi }{4}{\rm{ }} = {\rm{ }}\pi {\rm{ - }}\frac{\pi }{4} + k2\pi ;k \in Z\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = {\rm{ }}k2\pi ;k \in Z\\x{\rm{ }} = {\rm{ }}\frac{\pi }{2} + k2\pi ;k \in Z\end{array} \right.\end{array}\)

Mà \(x \in \left[ {0;\pi } \right]\) nên \(x \in \left\{ {0;\frac{\pi }{2}} \right\}\)

Vậy phương trình đã cho có số nghiệm là 2.

Chọn C