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

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Question 445384: How do you solve cos2Θ = cosΘ where Θ, 0° ≤ Θ < 360°
Answer by poliphob3.14(115) About Me  (Show Source):
You can put this solution on YOUR website!
For convenience substitute theta=x and modify the trigonometric equation:
cos2x=cosx, since cos2x=%28cosx%29%5E2-%28sinx%29%5E2 and
%28sinx%29%5E2=1-%28cosx%29%5E2, substituting we get:%28cosx%29%5E2-%28sinx%29%5E2=cosx =>
%28cosx%29%5E2-1%2B%28cosx%29%5E2=cosx => 2%28cosx%29%5E2-cosx-1=0. In the last equation
substitute cosx=y and get the quadratic equation: 2y%5E2-y-1=0,
Solving this equation 2y%5E2-y-1=%282y%2B1%29%28y-1%29=0 we find y=1 and y=-1/2.
Trigonometric equation cos2x=cosx is equivalent with two new equations:
cosx=-1 and cosx=-1%2F2. Solving these equations we have;
cosx=-1 => x=180degree and cosx=-1%2F2 => x=120, and
x=240degree.