Question

How do you solve `cos2x = sinx + cosx` on the interval [0,2pi]?

Answer 1

`x = 0,135,315,90,180`

Explanation:

`cos 2x = sin x + cos x`

`Cos^2 x - sin^2 x = cos x + sin x`

`(cosx + sinx)(cosx - sinx) = cosx + sin x`

`cancel((cosx + sinx))(cosx - sinx) =cancel( (cosx + sin x))`

`cos x - sin x = 1`

Square both sides

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

`cos ^2 x + sin ^2 x -2sinx cosx = 1`

`cancel1 - 2sinx cos x = cancel1`

`sin 2x = 0`

`=> 2x = 0=> x = 0`

Another method

`cos 2x - sin x - cos x = 0`

`cos^2x - sin^2x -(sinx + cosx) = 0`

`(Cosx - sinx)(cosx + sinx) -(sinx + cosx) = 0`

`color(blue)((sinx + cosx))color(red)((cosx - sinx -1)) = 0`

`color(blue)("Blue part")`

`color(blue)((sinx + cosx)) = 0`

`color(blue)(sinx = - cos x)`

`tanx = -1`

`x = 135,315`

`color(red)("red part")`

`color(red)((cosx - sinx -1)) = 0`

`color(red)(cos x - sin x = 1`

Square both sides

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

`cos ^2 x + sin ^2 x -2sinx cosx = 1`

`cancel1 - 2sinx cos x = cancel1`

`sin 2x = 0`

`=> 2x = 0,180,360=> x = 0,90,180`

Answer 2

`0, 5pi/4,, 3pi/2, 7pi/4 and 2pi`..

Explanation:

cos 2x = `cos^2x-sin^2x` = `(cos x - sin x)(cos x + sin x)`

The given equation is
`(cos x - sin x-1)(cos x + sin x) = 0`

Solve separately `cos x - sin x-1 = 0` and `cos x + sin x = 0`.

The solutions for the first are x = 0, `3pi/2, 2pi`.
For the other, x = `3pi/4, 7pi/4`.