# pre calc

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

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1. 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.

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2. 👎
2. 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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2. 👎

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