Solve `(x-1)|x+1|cosx>0,forx in [-pi,pi]`

Solve sin^(-1) (sin 6x) = x, x in [0,pi]

If f(x)=sqrt(1-sin2x) , then f^(prime)(x) is equal to (a) -(cosx+sinx) ,for x in (pi/4,pi/2) (b) cosx+sinx ,for x in (0,pi/4) (c) -(cosx+sinx) ,for x in (0,pi/4) (d) cosx-sinx ,for x in (pi/4,pi/2)

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

Differentiate cos^(-1)(cosx),\ \ x in [0,\ 2pi]

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

Solve cos^(-1)(cosx)>sin^(-1)(sinx),x in [0,2pi]

tan^(-1)[(cosx)/(1+sinx)] is equal to pi/4-x/2,forx in (-pi/2,(3pi)/2) pi/4-x/2,forx in (-pi/2,pi/2) pi/4-x/2,forx in (-pi/2,pi/2) pi/4-x/2,forx in (-(3pi)/2,pi/2)

Differentiate the following function with respect to x : sin^(-1)(sinx),x in [0,2pi] cos^(-1)(cosx),x in [0,2pi] tan^(-1)(tanx),x in [0,pi]-{pi/2}

Let f:[0,(pi)/(2)]toR be continuous and satisfy f'(x)=(1)/(1+cosx) for all x in(0,(pi)/(2)) . If f(0)=3 then f((pi)/(2)) has the value equal to :

Solve sin x tan x -sin x+ tan x-1=0 for x in [0, 2pi] .