Question

$2sin^{2} ( pi +x)-cos( -frac{ pi }{2} -x)=0$

на промежутке [ - frac{5 pi }{2} ; - pi ][/tex]

Answer 1

2sin²(π+x) -cos(-π/2-x) =0 ; 
2( -sinx)² - cos( -(π/2+x)) =0  ;
2sin²x -cos(π/2+x) =0 ;
2sin²x +sinx =0 ;
2sinx(sinx +1/2) =0 ;
[sinx =0 ; sinx = -1/2 .
[x=π*k ; x = (-1)^(k+1) π/6+π*k , k∈Z.