Question

How do you find all solutions to the equation in the interval [0,2pi] given `cos2x - sinx = 0`?

Answer 1

In the interval `[0,2pi]` Solution: `x=30^0,150^0,270^0`

Explanation:

`cos2x-sinx=0 or 1-2sin^2x-sinx=0 (cos2x=1-2sin^2x) or 2sin^2x+sinx-1=0 or (2sinx-1)(sinx+1)=0 :. either (2sinx=1 ; sinx=1/2) or sinx =-1` When `sinx =-1; x= 270^0`. When `sinx=1/2; x=30^0 , 150^0`[Ans]