Question

How do you solve for the exact solutions in the interval [0,2pi] of `cosx=sin2x`?

Answer 1

`x=pi/2,(3pi)/2 |x=pi/6,(5pi)/6`

Explanation:

`cosx=sin2x`

`=>cosx=2sinxcosx`[since`sin2x=2sinxcosx`]

`=>2sinxcosx-cosx=0`

`=>cosx(2sinx-1)=0`

`cosx=0 |sinx=1/2`

`x=pi/2,(3pi)/2 |x=pi/6,(5pi)/6`