Question

Срочно!!!Решите пожалуйста!!!!!!
sin^2 x-cos2x=2-2sin2x

cos(x+90°)*cos(x+30°)=1/2

Answer 1

Ответ:

Объяснение:

$sin^2(x) - cos2x = 2 - 2sin2x\\sin^2x - cos^2x + sin^2x = 2(sin^2x+cos^2x) - 4sinxcosx\\3cos^2x = 4sinxcosx\\cosx(3cosx - 4sinx) = 0\\cosx = 0 => x = \frac{\pi }{2} + \pi n\\4sinx = 3cosx => tgx = 3/4 => x = arctg(3/4) + \pi m$

n,m ⊆ Z

$cos(x+\frac{\pi }{2} ) * cos(x+\frac{\pi }{6}) = 1/2\\ -sinx * (cosxcos\frac{\pi}{6} - sinxsin\frac{\pi}{6}) = 1/2\\-sinx*(\frac{\sqrt{3}}{2}cosx - \frac{1}{2} sinx) = 1/2 | * 2\\-sinx*(\sqrt{3}cosx - sinx) = 1 = sin^2x+cos^2x\\sin^2x - \sqrt{3}sinxcosx - sin^2x -cos^2x = 0\\cos^2x + \sqrt{3}sinxcosx = 0\\cosx(cosx + \sqrt{3}sinx) = 0\\cosx = 0 => x = \frac{\pi}{2} + \pi n\\sinx = -\frac{\sqrt{3}}{3} cosx => tgx = -\frac{\sqrt{3}}{3} => x = -\frac{\pi}{6} + \pi m$

n,m ⊆ Z