Question

Помогите! Нужна помощь

Answer 1

$\displaystyle 2 \sin^2x+\sqrt{3} \sin2x =1+\cos2x\\2 \sin^2x+\sqrt{3} \sin2x = \sin^2x+\cos^2x+\cos^2x-\sin^2x\\2 \sin^2x+\sqrt{3} \sin2x = 2 \cos^2x\\2 \sin^2x + \sqrt{3} \cdot 2 \sin x \cos x=2 \cos^2x \;\; |\div 2\cos^2x\\tg^2x +\sqrt{3} \;tgx=1\\tg^2x+\sqrt{3} \; tgx-1=0\\D=(\sqrt{3})^2-4 \cdot 1 \cdot (-1)=7\\tgx_1=\frac{-\sqrt{3} +\sqrt{7} }{2} \\tgx_2=\frac{-\sqrt{3} -\sqrt{7} }{2}$

$\displaystyle \left [ {{x_1=arctg\big(\frac{-\sqrt{3} +\sqrt{7} }{2}\big)+\pi n; \; n\in Z} \atop {x_2=arctg\big(\frac{-\sqrt{3} -\sqrt{7} }{2}\big)+\pi n; \; n\in Z}} \right.$