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2 cos2(x) - 5 cos(x) - 3 = 0

To save some writing, let's temporarily write "Q" instead of "cos(x)".

2 Q2 - 5 Q - 3 = 0

Factor the trinomial:

(2 Q + 1) (Q - 3) = 0

2Q + 1 = 0

2Q = -1

Q = - 1/2

Q - 3 = 0

Q = 3

Recalling that 'Q' is a cosine of an angle, (Q = -1/2) is the root that makes sense,

because the cosine of a real angle can never be 3 .

So now, we can get rid of 'Q' and go back to the cosine:

cos(x) = -1/2

x = 120 degrees

x = 240 degrees

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Q: How would you solve the following over the interval 0 is less than or equal to x which is less than 2pi here's the problem 2cos squared x minus 5cos x minus 3 equals 0?
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