Question

sin^2x=0,75 на интервале -2п -п/2
Уравнение

Answer 1

$sin^2(x)=0,75<=>sin^2(x)=\frac{3}{4}\\sin(x)=\frac{\sqrt{3}}{2}=>x=\frac{\pi}{3}+2\pi k,x=\frac{2\pi}{3}+2\pi k\\sin(x)=-\frac{\sqrt{3}}{2}=>x=\frac{4\pi}{3}+2\pi k,x=\frac{5\pi}{3}+2\pi k$