2.

\(sin3x+cos2x=1+2sinx.cos2x\)

\(\Leftrightarrow sin3x+cos2x=1+sin3x-sinx\)

\(\Leftrightarrow cos2x+sinx-1=0\)

\(\Leftrightarrow-2sin^2x+sinx=0\)

\(\Leftrightarrow\left[{}\begin{matrix}sinx=0\\sinx=\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=k\pi\\x=\dfrac{\pi}{6}+k2\pi\\x=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)

1.

\(cos3x-cos4x+cos5x=0\)

\(\Leftrightarrow cos3x+cos5x-cos4x=0\)

\(\Leftrightarrow2cos4x.cosx-cos4x=0\)

\(\Leftrightarrow\left(2cosx-1\right)cos4x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}cosx=\dfrac{1}{2}\\cos4x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\4x=\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\pi}{3}+k2\pi\\x=\dfrac{\pi}{8}+\dfrac{k\pi}{4}\end{matrix}\right.\)

3.

\(cos2x-cosx=2sin^2\dfrac{3x}{2}\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.sin\dfrac{x}{2}+2sin^2\dfrac{3x}{2}=0\)

\(\Leftrightarrow2sin\dfrac{3x}{2}.\left(sin\dfrac{x}{2}+sin\dfrac{3x}{2}\right)=0\)

\(\Leftrightarrow sin\dfrac{3x}{2}.sinx.cos\dfrac{x}{2}=0\)

Đến đây dễ rồi tự làm tiếp nha.

\(\Leftrightarrow2\cos^3x+2\cos^2x-1+\left(1-\cos^2x\right)=0\)

\(\Leftrightarrow2\cos^3x+\cos^2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=\dfrac{-1}{2}\\\cos=0\end{matrix}\right.\)

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

\(pt\Leftrightarrow2cos\left(\frac{11x}{8}\right)cos\left(\frac{5x}{8}\right)=2\)

\(\Leftrightarrow cos\left(\frac{11x}{8}\right)cos\left(\frac{5x}{8}\right)=1\)

\(\Leftrightarrow cos\left(\frac{11x}{8}\right)=1\) và \(cos\left(\frac{5x}{8}\right)=1\) hoặc \(cos\left(\frac{11x}{8}\right)=-1\) và \(cos\left(\frac{5x}{8}\right)=-1\)

Đến đây dễ r`

\(3cos2x+4cos^3x-cos3x=0\\ \Leftrightarrow3cos2x+4cos^3x-\left(4cos^3x-3cosx\right)=0\\ \Leftrightarrow cos2x=cos\left(\pi-x\right)\\ \Leftrightarrow\left[{}\begin{matrix}2x=\pi-x+k2\pi\\2x=x-\pi+k2\pi\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{3}+\frac{k2\pi}{3}\\x=-\pi+k2\pi\end{matrix}\right.\\ \Leftrightarrow x=\frac{\pi}{3}+\frac{k2\pi}{3}\)

bạn ơi, sao cos3x chuyển thành 4cos³x-3cosx vậy ạ

\(\Leftrightarrow2^{\cos2x-1}\left(2\cos x-1\right)=2\cos^2x\left(2\cos x-1\right)\)

\(\Leftrightarrow\left(2\cos x-1\right)\left(2^{\cos2x}-2\cos^2x\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}\cos x=\frac{1}{2}\\2^{\cos2x}=\cos2x+1\end{array}\right.\)

* Với \(\cos x=\frac{1}{2}\) ta có \(x=\frac{\pi}{3}=k2\pi,k\in Z\)

* Với \(2^{\cos2x}=\cos2x+1\) (*), đặt \(t=\cos2x;t\in\left[-1;1\right]\)

Phương trình trở thành \(2^t-t-1=0\)

Xét hàm số \(f\left(t\right)=2^t-t-1,t\in\left[-1;1\right]\)

Có \(f'\left(t\right)=2^t\ln2-1,t\in\left[-1;1\right];f'\left(t\right)=0\) có đúng 1 nghiệm  nên phương trình \(f\left(t\right)=0\) có tối đa 2 nghiệm. Mà \(f\left(0\right)=f\left(1\right)=0\) nên \(t=0;t=1\) là tất cả các nghiệm của phương trình \(f\left(t\right)=0\)

Do đó phương trình (*) \(\Leftrightarrow\left[\begin{array}{nghiempt}\cos2x=0\\\cos2x=1\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{\pi}{4}+k\frac{\pi}{2}\\x=k\pi\end{array}\right.\) \(k\in Z\)

Vậy phương trình đã cho có 3 nghiệm là :

\(x=\frac{\pi}{3}+k2\pi;x=\frac{\pi}{4}+k\frac{\pi}{2};x=k\pi;k\in Z\)

1.

        \(\cos2x+\sin\left(x+\frac{pi}{4}\right)=0\)

\(\Leftrightarrow\sin\left(x+\frac{pi}{4}\right)=-\cos2x\)

\(\Leftrightarrow\sin\left(x+\frac{pi}{4}\right)=\sin\left(2x-\frac{pi}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{pi}{4}=2x-\frac{pi}{2}+k2pi\\x+\frac{pi}{4}=pi-2x+\frac{pi}{2}+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{3}{4}pi+k2pi\\3x=+\frac{5}{4}pi+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}pi+k2pi\\x=\frac{5}{12}pi+k\frac{2}{3}pi\end{cases}}\)

2.

\(\sin\left(3x-\frac{5pi}{6}\right)+\cos\left(3x+\frac{3pi}{6}\right)=0\)

\(\Leftrightarrow\sin\left(3x-\frac{5pi}{6}\right)=-\cos\left(3x+\frac{3pi}{6}\right)\)

\(\Leftrightarrow\sin\left(3x-\frac{5pi}{6}\right)=\sin\left(3x+\frac{3pi}{6}-\frac{pi}{2}\right)\)

\(\Leftrightarrow\orbr{\begin{cases}3x-\frac{5pi}{6}=3x+\frac{3pi}{6}-\frac{pi}{2}+k2pi\\3x-\frac{5pi}{6}=pi-3x-\frac{3pi}{6}+\frac{pi}{2}+k2pi\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0x=\frac{5pi}{6}+k2pi\left(VN\right)\\6x=\frac{11pi}{6}+k2pi\end{cases}}\)

\(\Leftrightarrow x=\frac{11pi}{36}+k\frac{1}{3}pi\)

a, \(\dfrac{1+cosx+cos2x+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{1+cos2x+cosx+cos3x}{2cos^2x+cosx-1}\)

\(=\dfrac{2cos^2x+2cos2x.cosx}{cos2x+cosx}\)

\(=\dfrac{2cosx\left(cos2x+cosx\right)}{cos2x+cosx}=2cosx\)

b) \(cos\dfrac{5x}{2}.cos\dfrac{3x}{2}+sin\dfrac{7x}{2}.sin\dfrac{x}{2}\)

\(=cos\dfrac{4x+x}{2}.cos\dfrac{4x-x}{2}+sin\dfrac{4x+3x}{2}.sin\dfrac{4x-3x}{2}\)

\(=\dfrac{1}{2}\left(cos4x+cosx\right)-\dfrac{1}{2}\left(cos4x-cos3x\right)\)

\(=\dfrac{1}{2}\left(cosx+cos3x\right)=\dfrac{1}{2}.2cos2x.cos\left(-x\right)\)\(=cosx.cos2x\)