Question

2sin²x +5cosx -4=0 ПОМОГИТЕ !!!!!!

Answer 1

$\displaystyle 2sin^2x+5cosx-4=0\\\\2(1-cos^2x)+5cosx-4=0\\\\2-2cos^2x+5cosx-4=0\\\\-2cos^2x+5cosx-2=0\\\\2cos^2x-5cosx+2=0$

$\displaystyle cosx=t\\\\2t^2-5t+2=0\\\\D=25-16=9\\\\t_{1.2}= \frac{5\pm 3}{4}\\\\t_1=2; t_2= \frac{1}{2}\\\\cosx \neq 2\\\\cosx= \frac{1}{2}\\\\x=\pm \frac{ \pi}{3}+2 \pi n; n\in Z$