Question

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Answer 1

$2\cos^2(3x)+\sin(5x)=1 \medskip \\ \left(2\cos^2(3x)-1\right)+\sin(5x)=0 \medskip \\ \cos(6x)+\sin(5x)=0 \medskip \\ \sin\left(\dfrac{\pi}{2}-6x\right)+\sin(5x)=0 \medskip \\ 2\sin\left(\dfrac{\frac{\pi}{2}-6x+5x}{2}\right)\cos\left(\dfrac{\frac{\pi}{2}-6x-5x}{2}\right)=0 \medskip \\ \sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)\cos\left(\dfrac{11x}{2}-\dfrac{\pi}{4}\right)=0$

$1)~\sin\left(\dfrac{x}{2}-\dfrac{\pi}{4}\right)=0 \medskip \\ \dfrac{x}{2}-\dfrac{\pi}{4}=\pi m,~m\in\mathbb{Z} \medskip \\ x=\dfrac{\pi}{2}+2\pi m,~m\in\mathbb{Z} \medskip \\ 2)~\cos\left(\dfrac{11x}{2}-\dfrac{\pi}{4}\right)=0 \medskip \\ \dfrac{11x}{2}-\dfrac{\pi}{4}=\dfrac{\pi}{2}+\pi k,~k\in\mathbb{Z} \medskip \\ x=\dfrac{3\pi}{22}+\dfrac{2\pi k}{11},~k\in\mathbb{Z}$

Ответ. $x=\dfrac{\pi}{2}+2\pi m,~m\in\mathbb{Z};~x=\dfrac{3\pi}{22}+\dfrac{2\pi k}{11},~k\in\mathbb{Z}$