Question

помогите срочно надо

Answer 1

Решение во вложении)

Answer 2

$sin2x=\sqrt3cosx\\2sinx*cosx-\sqrt3cosx=0\\cosx(2sinx-\sqrt3)=0\\\\ \left[\begin{array}{ccc}cosx=0\\2sinx-\sqrt{3}=0\end{array}\right=\ \textgreater \ \left[\begin{array}{ccc}x=\frac{\pi}2+\pi n;n\in Z\\sinx=\frac{3}2\end{array}\right=\ \textgreater \ \\\\\left[\begin{array}{ccc}x=\frac{\pi}2+\pi n;n\in Z\\x=\frac{\pi}3+2\pi n;n\in Z\\x=\frac{2\pi}3+2\pi n;n\in Z\end{array}\right$