Question

2*cos2x=sin(3*pi/2-x)-2

Answer 1

по формуле приведения sin(3pi/2-x)=-cosx
2cosx=-cosx-2
3cosx=-2
cosx=-2/3
x=+-arcos(-2/3)

Answer 2

$2cos2x=sin(frac{3pi}{2}-x)-2;\ 4cos^2x-2=sinfrac{3pi}{2}cos x-cosfrac{3pi}{2}sin x-2;\ 4cos^2x-2=-cos x-2;\ 4cos^2x+cos x=0;\ t=cos x; -1leq t leq1;\ 4t^2+t=0;\ t(4t+1)=0;\ t_1=0;\ t_2=-frac{1}{4};\ cos x =0; x_1=frac{pi}{2}+pi n nin Z;\ cos x=-frac{1}{4}; x=pmarccos(-frac{1}{4})+2 pi k, kin Z$