Question

Решите уравнение: а) sin3x-√3cos2x=sinx. б) 2sin2x-sin^2x=3cos^2x

Answer 1

sin3x - sinx = √3cos2x

2sinxcos2x - √3cos2x = 0

cos2x (2sinx - √3) = 0

2x = p/2 + 2pn

x = p/4 + pn

sinx = √3/2

x = (-1)^n * p/6 + pn

б) 3cos^2x + sin^2x - 2sin2x  = 0

3 - 3sin^2x + sin^2x - 2sin2x = 0

-sin^2x - 2sin2x + 3 = 0

sin^2x + 2sinxcosx + 3 = 0

tg^2x + 2tgx + 3 + 3tg^2x = 0

4tg^2x + 2tgx + 3 = 0