Question

2cosx=1-2(cos x)^2-sqrt3*sin(2x)
я так понимаю что надо решать методом доп аргумента но что то не сходиться.

Answer 1

2cosx=1-2cos²x-√3sin2x
2cosx=-cos2x-√3sin2x
cosx=-1/2*cos2x-√3/2*sin2x
cosx=-cos2x*cosπ/3-sin2x*sinπ/3
cosx=-cos(2x-π/3)
cos(2x-π/3)+cosx=0
2cos(3x/2-π/6)*cos(x/2-π/6)=0
cos(3x/2-π/6)=0⇒3x/2-π/6=π/2+πk⇒3x/2=2π/3+πk⇒x=4π/9+2πk/3,k∈z
cos(x/2-π/6)=0⇒x/2-π/6=π/2+πk⇒x/2=π/3+πk⇒x=4π/3+2πk,k∈z

Answer 2

$2cosx=1-2cos^2x- \sqrt3sin2x\\ 2cosx=-cos2x-\sqrt3sin2x\\ cosx=- \frac{1}{2} cos2x-\frac{\sqrt3}{2}sin2x\\ cosx=- cos\frac{\pi}{3} cos2x-sin \frac{\pi}{3}sin2x\\ cosx=-cos(2x-\frac{\pi}{3})$
$cosx+cos(2x-\frac{\pi}{3})=0\\ 2cos\dfrac{x+2x-\frac{\pi}{3}}{2}cos\dfrac{x-2x+\frac{\pi}{3}}{2}=0\\ cos( \frac{3x}{2}- \frac{\pi}{6})cos( -\frac{x}{2}+ \frac{\pi}{6})=0$
$cos( \frac{3x}{2}- \frac{\pi}{6})=0$ или $cos( \frac{x}{2}- \frac{\pi}{6})=0$
$\frac{3x}{2}- \frac{\pi}{6}= \frac{\pi}{2}+ \pi k$ или $\frac{x}{2}- \frac{\pi}{6}= \frac{\pi}{2}+ \pi n$
9x - π = 3π + 6πk или 3x - π = 3π + 6πn
$x= \frac{4\pi}{9} + \frac{2\pi k}{3} , k \in Z$ или $x= \frac{4\pi}{3} + 2\pi n , n \in Z$
Ответ: $\frac{4\pi}{9} + \frac{2\pi k}{3}; \frac{4\pi}{3} + 2\pi n;\ k, n \in Z.$