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Question
Solve the following equations:
2cos 2x – 7 cos x + 3 = 0
Solution
2 cos2x – 7 cos x + 3 = 0
2 cos2x – 6 cos x – cos x + 3 = 0
2 cos x (cos x – 3) – 1(cos x – 3) = 0
(2 cos x – 1)(cos x – 3) = 0
2 cos x – 1 = 0 or cos x – 3 = 0
cos x = `1/2` or cos x = 3|
Since – 1 ≤ cos x ≤ 1, we have
cos x = 3 is not possible.
∴ cos x = `1/2`
cos x = `cos pi/3`
The general solution is x = `2"n"pi +- pi/3`, n ∈ Z
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