Solve Trig Equation: cos 2x - cos^2x = 0

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The equation cos 2x - cos^2x = 0 simplifies to cos^2x - 1 = 0, leading to solutions where cos x equals -1 and 1. The calculated values of x are 180° and 0, but since the interval is -180° < x < 180°, neither 180° nor -180° can be included as solutions. The only valid solution within the specified range is x = 0, while including endpoints would allow for additional solutions at -π and π. The discussion emphasizes the importance of checking the interval constraints when determining valid solutions.
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soo okay...i have this question that I am stuck on...

Solve for x,

cos 2x - cos^2x = 0, -180degrees<x<180degrees

i tried solving it...

cos 2x - cos^2x = 0
[2cos^2x -1]-cos^2x = 0
cos^2x -1 = 0 <--(?)
(cos x -1)(cos x +1) = 0
therefore... 1)cos x = -1 , 2)cos x = 1

so...

1)cos x = -1
x = cos^-1 (-1)
x= 180

2) cos x = 1
x = 0

but i have tried on my calculator...and -180 degrees can be one of the answers too...i don't get how to get -180

SOMEONE HELPP...and reply FASTT please and Thank yOU in advance
 
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Well where is cos(x) equal to -1 besides x=180°?

(Also, since the original problem asks for solutions greater than -180° and less than 180°, then neither 180° nor -180° would be solutions. Maybe the problem had -180° ≤ x ≤ 180°)
 
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the thing is...i don't even know if i did the equation right... T.T it seems wrong
 
How does it "seem wrong"? Either you made a mistake and it's wrong or you didn't and it's right.

I didn't see any mistakes except for the two things I pointed out.

You can plug in the values of x you found into the original equation to check your work.
 
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umm..thanks..
 
You're doing all right, the solution are k\pi with k \in \mathbb{Z}.
The only solution in the given interval is x = 0, if the inequalities weren't strict, you have to add pi and -pi as solutions.
 

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