Question

Помогите, срочно, пж!!!
Решить уравнение​

Answer 1

$\displaystyle (4sin^2x-sin2x-3)*\sqrt{cosx}=0\\\\ODZ: cosx\geq 0; \\\\-\frac{\pi }{2}+2\pi n\leq x\leq \frac{\pi }{2}+2\pi n; n \in Z$

$\displaystyle cos x=0; 4sinx^2x-sin2x-3=0\\\\cosx=0; x=\frac{\pi }{2}+\pi n; n \in Z \\\\4sin^2x-2sinx*cosx-3(sin^2x+cos^2x)=0\\\\sin^2x-2sinx*cosx-3cos^2x=0|:cos^2x\\\\tg^2x-2tgx-3=0\\\\D=4+12=16\\\\tgx=\frac{2 \pm 4}{2} \\\\ tgx=3; tgx=-1$

с учетом ОДЗ:

$\displaystyle x=arcTg 3+ 2\pi n; n \in Z\\\\ x= -\frac{\pi }{4}+2\pi n; n \in Z$

овтет:

$\displaystyle \frac{\pi }{2}+\pi n; -\frac{\pi }{4}+2\pi n; arctg(3)+2\pi n$