Question

2sinв степени2x-3sinx+1=0

Answer 1

$2sin^2x-3sinx+1=0\\sinx=t,t\in[-1;1]\\2t^2-3t+1=0\\D=9-4*2=9-8=1\\\\t_1=\frac{3+1}4=\frac 44=1\\\\t_2=\frac{3-1}4=\frac{2}4=\frac{1}2\\\\ \left[\begin{array}{ccc}sinx=1\\sinx=\frac{1}2\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}x=\frac{\pi}2+2\pi n;n\in Z\\x=\frac{\pi}6+2\pi n;n\in Z\\x=\frac{5\pi}6+2\pi n;n\in Z\end{array}\right$