а) По формуле приведения: cos ( 3 π 2 + x ) = sin x \cos\left(\frac{3\pi}{2}+x\right)=\sin x cos ( 2 3 π + x ) = sin x 2 sin 2 x + 3 2 sin x + 2 = 0 ⇔ [ sin x = − 2 2 sin x = − 2 2\sin^2x+3\sqrt{2}\sin x+2=0\Leftrightarrow \displaystyle \left[ \begin{array}{l l} \sin x=-\frac{\sqrt{2}}{2}\\ \sin x=-\sqrt{2} \end{array} \right. 2 sin 2 x + 3 2 sin x + 2 = 0 ⇔ [ sin x = − 2 2 sin x = − 2 sin x = − 2 \sin x=-\sqrt{2} sin x = − 2 sin x ∈ [ − 1 ; 1 ] \sin x\in [-1;1] sin x ∈ [ − 1 ; 1 ] sin x = − 2 2 ⇔ [ x = − π 4 + 2 π k x = − 3 π 4 + 2 π k , k ∈ Z \displaystyle \sin x=-\frac{\sqrt{2}}{2} \Leftrightarrow \left[ \begin{array}{l l} x=-\frac{\pi }{4} +2\pi k\\ x=-\frac{3\pi }{4} +2\pi k \end{array} \right. ,k\in Z sin x = − 2 2 ⇔ [ x = − 4 π + 2 π k x = − 4 3 π + 2 π k , k ∈ Z [ 5 π 2 ; 4 π ] \left[\frac{5\pi}{2};4\pi\right] [ 2 5 π ; 4 π ]
Нам подходят корни: 3 π + π 4 = 13 π 4 4 π − π 4 = 15 π 4 3\pi+\frac{\pi}{4}=\frac{13\pi}{4}\\ 4\pi-\frac{\pi}{4}=\frac{15\pi}{4} 3 π + 4 π = 4 13 π 4 π − 4 π = 4 15 π − π 4 + 2 π k ; − 3 π 4 + 2 π k : k ∈ Z -\frac{\pi}{4}+2\pi k;-\frac{3\pi}{4}+2\pi k:k\in Z − 4 π + 2 π k ; − 4 3 π + 2 π k : k ∈ Z 13 π 4 ; 15 π 4 \frac{13\pi}{4};\frac{15\pi}{4} 4 13 π ; 4 15 π
