SOLUTION: How do you solve: where &#920;, 0° &#8804; &#920; < 360° cos2&#920; = cos&#920;

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Question 445380: How do you solve: where Θ, 0° ≤ Θ < 360°
cos2Θ = cosΘ
Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
cos%282THETA%29+=+cos%28THETA%29
Using the identity cos%282THETA%29+=+2%28cos%28THETA%29%29%5E2+-+1 we have

This is a quadratic in cos%28THETA%29. Let x = cos%28THETA%29. Then we can write:
2x%5E2+-+x+-+1+=+0
Solve for x using the quadratic formula:
x+=+%281+%2B-+sqrt%281+%2B+8%29%29%2F4
This gives x = 1, -1/2
Since x+=+cos%28THETA%29 we need to find values for THETA which give cos%28THETA%29+=+1, -1%2F2
The inv. cos of 1 = 0 deg.
The inv. cos of -1/2 = 120 deg.
Since the THETA ranges from 0 to 360 deg. and since 2THETA = 240 deg., we need to include 240 deg.
So the answers are THETA+=+0deg, THETA+=+120deg, THETA+=+240deg