Question

How do you solve `cos2x = sinx` on the interval `0 <= x<= 2pi`?

Answer 1

If `cos(2x) = sin(x)`
then
`1-2sin^2(x) = sin(x)`

`2sin^2(x) +sin(x) -1 =0`

Substituting `k=sin(x)`
`2k^2+k-1 = 0`

`(2k-1)(k+1) = 0`

`sin(x) = 1/2` or `sin(x) =-1`

If `sin(x) = 1/2` (for `0<=x<=2pi`)
`x=pi/6 =30^o` or `x= (5pi)/6 = 150^o`

If `sin(x) = -1` (for `0<=x<=2pi`)
`x=(3pi)/2 = 270^o`

So `x epsilon {pi/6, (5pi)/6, (3pi)/2}`
(or their equivalent in degrees)