Question

How do you solve `2-2cos^2x=sinx+1` for `0<=x<=2pi`?

Answer 1

Solution: `x=90^0,210^0,330^0`

Explanation:

`2-2cos^2x=sinx+1 or 2(1-cos^2x) -sinx-1=0 or 2sin^2x-sinx-1=0 or 2sin^2x-2sinx+sinx-1=0 or 2sinx(sinx-1)+1(sinx-1)=0or (2sinx+1)(sinx-1)=0:.2sinx=-1 or sinx=1 :. sinx=-1/2 or sinx=1` when `sinx= -1/2; x=210^0 and x=330^0` when `sinx=1 ; x=90^0 :.x=90^0,210^0,330^0`[Ans]