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

2 Answers
Mar 17, 2016

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

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.