Ta có: \(2\cos 3x(2\cos 2x + 1) = 1\)

=>\(4\cos3x\cos2x+2\cos3x=1\)

=>\(2(\cos 5x + \cos x) + 2\cos 3x = 1\)

=>\(2\cos5x+2\cos3x+2\cos x=1\) (1)

TH1: sin x=0

=>\(x=k\pi\)

Thay vào (1), ta được: \(2\cdot cos\left(5k\pi\right)+2\cdot cos\left(3k\pi\right)+2\cdot cos\left(k\pi\right)=1\)

=>2*\(\pm1\) (2*1+1)=1

=>\(\pm6\)=1(vô lý)

=>Loại

TH2: sin x<>0

(1)=>\(2\sin x \cdot \cos 5x + 2\sin x \cdot \cos 3x + 2\sin x \cdot \cos x = \sin x\)

=>\((\sin6x-\sin4x)+(\sin4x-\sin2x)+\sin2x=\sin x\)

=>sin 6x=sin x

=>\(\left[\begin{array}{l}6x=x+k2\pi\\ 6x=\pi-x+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=k2\pi\\ 7x=\pi+k2\pi\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac{k2\pi}{5}\\ x=\frac{\pi}{7}+\frac{k2\pi}{7}\end{array}\right.\)

TH1: \(x=\frac{k2\pi}{5}\)

\(x\in\left\lbrack-4\pi;6\pi\right\rbrack\)

=>\(-4\pi\le\frac{2k\pi}{5}\le6\pi\)

=>-10<=k<=15

Ta sẽ loại các giá trị k thỏa mãn \(x=k\pi\)

=>k∉{-10;-5;0;5;10;15}

\(S_1 = \sum_{k=-10}^{15} \frac{2k\pi}{5} - \sum_{k \in \{-10,-5,0,5,10,15\}} \frac{2k\pi}{5}\)

\(=\frac{2\pi}{5}\cdot\frac{(-10 + 15) \cdot26}{2}-\frac{2\pi}{5}\cdot15=20\pi\)

TH2: \(x=\frac{\pi+k2\pi}{7}\)

\(x \in [-4\pi; 6\pi]\)

=>\(-4\pi\le\frac{(2k+1)\pi}{7}\le6\pi\)

=>\(-29\le2k+1\le42\)

=>-15<=k<=20

Ta sẽ loại các giá trị k khiến cho x=kΩ

=>k∉{-11;-4;3;10;17}

\(S_2 = \sum_{k=-15}^{20} \frac{(2k+1)\pi}{7} - \sum_{k \in \{-11,-4,3,10,17\}} \frac{(2k+1)\pi}{7}\)

=>\(S_2 = \frac{216\pi}{7} - \frac{35\pi}{7} = \frac{181\pi}{7}\)

S=S1+S2

\(=20\pi+\frac{181\pi}{7}=\frac{321\pi}{7}\)

Đáp án C

A

A

=>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

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

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

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:

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

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

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

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

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

B

\(\Leftrightarrow2cos^2\left(x+\dfrac{pi}{3}\right)-1=0\)

=>\(cos\left(2x+\dfrac{2}{3}pi\right)=0\)

=>2x+2/3pi=pi/2+kpi

=>2x=-1/6pi+kpi

=>x=-1/12pi+kpi/2

mà \(x\in\left(-\dfrac{pi}{2};\dfrac{5}{6}pi\right)\)

nên \(x\in\left\{-\dfrac{1}{12}pi;\dfrac{5}{12}pi\right\}\)