Question

Доброго времени суток, помогите решить уравнение:
2cos²x = 1- (√3)*cos (3п/2+x)

Answer 1

$2\cos^{2} x = 1 - \sqrt{3} \cos \bigg(\dfrac{3\pi}{2} +x \bigg)\\2\cos^{2} x = \sin^{2}x + \cos^{2}x - \sqrt{3} \sin x\\\sin^{2}x - \cos^{2}x - \sqrt{3} \sin x = 0\\-\cos2x - \sqrt{3} \sin x = 0\\\cos2x + \sqrt{3} \sin x = 0\\1 - 2\sin^{2}x + \sqrt{3} \sin x = 0$

Замена: $\sin x = t, \ t \in [-1; \ 1]$

$1 - 2t^{2} + \sqrt{3} t = 0\\2t^{2} - \sqrt{3} t - 1 = 0\\D = 3 + 8 = 11\\x_{1} = \dfrac{\sqrt{3} - \sqrt{11}}{4} \\x_{2} = \dfrac{\sqrt{3}+\sqrt{11}}{4} > 1$

$\sin x = \dfrac{\sqrt{3} - \sqrt{11}}{4}\\x = (-1)^{n} \arcsin \bigg(\dfrac{\sqrt{3} - \sqrt{11}}{4} \bigg) + \pi n, \ n \in Z$

Ответ: $x = (-1)^{n} \arcsin \bigg(\dfrac{\sqrt{3} - \sqrt{11}}{4} \bigg) + \pi n, \ n \in Z$