Question

2sin^2x=V3cos (5П2 + x)   корни промежутка ( -П; П2)

Answer 1

2 sin^2 x = sgrt3 *( - sin x) = 0;
2 sin^2 x + sgrt3*sin x = 0;
sin x( 2sin x + sgrt3) = 0;
 1) sin x = 0; ⇒ x = pi k; k∈Z.
2) 2 sin x + sgrt3 =0;
 sin x = - sgrt3/2;⇒ x= ( -1) ^(k+1)*pi/3 + pik; k⇒Z.

[- pi; pi/2]
x1= - pi; x2= - pi/3 ; x3= 0.