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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?
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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