Question

The number of solutions of the equation

sin^-1 (cosx)=cos-1(sinx), x∈[π/2, 2π] is

Solution

sin^-1(cosx) = cos^-1(sinx)

cos^-1 (π/2-cosx)=cos^-1(sinx)

⇒ π/2-cosx = sinx

⇒ sinx + cosx = π/2

⇒(sinx+cosx)^2 = (π/2)^2

⇒sin^2x+cos^2x+2cosxsinx=π^2/4

⇒1+2cosxsinx=π^7/4

2cosxsinx=π^2/4-1

**cosx sinx = π^2-4/8 Ans **