Posted by **caitlyn** on Tuesday, April 26, 2011 at 9:14pm.

cos2x=cosx in interval of [0,2pie]

- pre calc -
**MathMate**, Tuesday, April 26, 2011 at 9:42pm
cos2x-cosx=0

expand the left hand side

cos²(x)-sin²(x)-cos(x)=0

2cos²(x)-1 - cos(x)=0

substitute c=cos(x)

2c²-c-1=0

c=(-1±√(9))/4

=-1 or 1/2

cos(x)=-1 when x=π (0≤x≤2π)

or

cos(x)=1/2 when x=π/3 or x=5π/3 (0≤x≤2π)

Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

- pre calc -
**MathMate**, Tuesday, April 26, 2011 at 9:57pm
Reiny is right.

There was a mistake in the solution of the quadratic.

c=(1±√(9))/4

=1 or -1/2

cos(x)=1 when x=0 or 2π (0≤x≤2π)

or

cos(x)=-1/2 when x=π±π/3 (0≤x≤2π)

Substitute each of the three solution into the original equation to make sure that the solutions are acceptable.

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