Question

How do you solve `cos^2x=1-sinx` between the interval `0<=x<=2pi`?

Answer 1

Solution; `x=0,x=pi/2,x=pi;x=2pi`. In interval `0<=x<=2pi`

Explanation:

`cos^2x=1-sinx or 1-sin^2x= 1-sinx or sin^2x-sinx=0 or sinx(sinx-1)=0 :.` either `sinx=0 or sinx-1=0 :. sinx=1` When `sinx=0; x=0,x=pi,x=2pi`. Since `sin0=0; sin pi=0 ; sin2pi=0`
When `sinx=1; x=pi/2`. Since `sin (pi/2)=0`
Solution; `x=0,x=pi/2,x=pi;x=2pi`. In interval `0<=x<=2pi`[Ans]