Question

3sin в квадрате x- 2sonx· cosx =3cos в квадрате x

Answer 1

3sin²x -2sinxcosx = 3cos²x
3sin²x -2sinxcosx - 3cos²x = 0
3($\frac{1-cos2x}{2}$) - sin2x -3($\frac{1+cos2x}{2}$) = 0
$\frac{3-3cos2x}{2}$ - sin2x - $\frac{3-3cos2x}{2}$ = 0
3-3cos2x - 2sin2x - 3 - 3cos2x = 0
-6cos2x - 2sin2x = 0
-6 - 2tg2x = 0
-2tg2x = 6
tg2x = -3
2x = arctg(-3) + πk,k∈z
x = -$\frac{arctg3}{2}$ + $\frac{ \pi k}{2}$, k∈z