Question

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

Answer 1

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

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