а) По формуле приведения: sin ( x + π ) = − sin x \sin(x+\pi)=-\sin x sin ( x + π ) = − sin x 4 sin x + 4 − sin x = 5 2 4^{\sin x}+4^{-\sin x}=\frac{5}{2} 4 s i n x + 4 − s i n x = 2 5 4 sin x = t , t > 0 4^{\sin x}=t,t>0 4 s i n x = t , t > 0 t + 1 t = 5 2 2 t 2 + 2 − 5 t 2 t = 0 ⇔ { [ t = 2 t = 1 2 t ≠ 0 [ t = 2 t = 1 2 ⇔ [ 4 sin x = 4 1 2 4 sin x = 4 − 1 2 ⇔ [ sin x = 1 2 sin x = − 1 2 ⇔ [ x = π 6 + 2 π k x = 5 π 6 + 2 π k x = − π 6 + 2 π k x = − 5 π 6 + 2 π k , k ∈ Z ⇔ ⇔ x = ± π 6 + π k , k ∈ Z t+\frac{1}{t}=\frac{5}{2}\\ \frac{2t^2+2-5t}{2t}=0\Leftrightarrow \displaystyle \begin{cases} \left[ \begin{array}{l l} t=2\\ t=\frac{1}{2} \end{array} \right.\\ t\neq 0 \end{cases}\\ \displaystyle \left[ \begin{array}{l l} t=2\\ t=\frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l l} 4^{\sin x} =4^{\frac{1}{2}}\\ 4^{\sin x} =4^{-\frac{1}{2}} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l l} \sin x=\frac{1}{2}\\ \sin x=-\frac{1}{2} \end{array} \right. \Leftrightarrow \left[ \begin{array}{l l} x=\frac{\pi }{6} +2\pi k\\ x=\frac{5\pi }{6} +2\pi k\\ x=-\frac{\pi }{6} +2\pi k\\ x=-\frac{5\pi }{6} +2\pi k \end{array} \right. ,k\in Z\Leftrightarrow \\ \displaystyle \Leftrightarrow x=\pm \frac{\pi }{6} +\pi k,k\in Z t + t 1 = 2 5 2 t 2 t 2 + 2 − 5 t = 0 ⇔ ⎩ ⎨ ⎧ [ t = 2 t = 2 1 t = 0 [ t = 2 t = 2 1 ⇔ [ 4 s i n x = 4 2 1 4 s i n x = 4 − 2 1 ⇔ [ sin x = 2 1 sin x = − 2 1 ⇔ x = 6 π + 2 π k x = 6 5 π + 2 π k x = − 6 π + 2 π k x = − 6 5 π + 2 π k , k ∈ Z ⇔ ⇔ x = ± 6 π + π k , k ∈ Z [ 5 π 2 ; 4 π ] \left[\frac{5\pi}{2};4\pi\right] [ 2 5 π ; 4 π ]
3 π − π 6 = 17 π 6 3 π + π 6 = 19 π 6 4 π − π 6 = 23 π 6 3\pi-\frac{\pi}{6}=\frac{17\pi}{6}\\ 3\pi+\frac{\pi}{6}=\frac{19\pi}{6}\\ 4\pi-\frac{\pi}{6}=\frac{23\pi}{6} 3 π − 6 π = 6 17 π 3 π + 6 π = 6 19 π 4 π − 6 π = 6 23 π ± π 6 + π k : k ∈ Z \pm\frac{\pi}{6}+\pi k:k\in Z ± 6 π + π k : k ∈ Z 17 π 6 ; 19 π 6 ; 23 π 6 \frac{17\pi}{6};\frac{19\pi}{6};\frac{23\pi}{6} 6 17 π ; 6 19 π ; 6 23 π
