Question

Решите уравнение 2синус^2 x/1-кос х=3

Answer 1

$\displaystyle \frac{2sin^2x}{1-cosx}=3\\\\ODZ: cosx\neq 1; x\neq 2\pi n; n \in Z\\\\2(1-cos^2x)=3(1-cosx)\\\\2-2cos^2x=3-3cosx\\\\2cos^2x-3cosx+1=0\\\\t=cosx; |t|\leq 1\\\\2t^2-3t+1=0\\\\D=9-8=1\\\\t_{1.2}=\frac{3 \pm 1}{4}\\\\ t_1=1; t_2=\frac{1}{2}$

но cosx≠1

тогда

$\displaystyle cosx=\frac{1}{2}\\\\x= \pm \frac{\pi }{3}+2\pi n; n \in Z$