Solve `cos2x >|sinx|,x in (pi/2,pi)`

Solve cos 2x=|sin x|, x in (-pi/2, pi) .

Solve cos2xgt|sinx|,x in(-(pi)/(2),pi)

Solve: cos 2x gt | sin x| x in [-pi,pi] .

If cos2x>|sin x| and x in(-(pi)/(2),pi) then x

82.Solve for x, sin x+cos x=(pi)/(2)

Solve |sinx +cos x |=|sinx|+|cosx|, x in [0,2pi] .

Solve the equation 2 (cos x+cos 2x)+sin 2x (1+2 cos x)=2 sin x for x in [-pi, pi] .

Solve 2 cos^(2) x+ sin x le 2 , where pi//2 le x le 3pi//2 .

If f(x)={(x+a sqrt(2) sinx"," ,0 lt x lt (pi)/(4)),(2x cotx+b",",(pi)/(4) le x le (pi)/(2)),(a cos 2x-b sinx",", (pi)/(2) lt x le pi):} is continuous at x=(pi)/(4) , then a - b is equal to

prove that cos^-1(sinx)= pi/2-x