How do you solve cos2x-cos^2x-2sinx+3 = 0 from [-2pi, pi/2]#?

1 Answer
Nghi N.
Jul 23, 2015

Solve f(x) = cos 2x - cos^2 x - 2sin x + 3 = 0
Ans: x = pi/2

Explanation:

Replace cos 2x and cos^2 x in terms of sin x. Call sin x = t

(1 - 2sin^2x) - (1 - sin^2 x) - 2sin x + 3 = 0
1 - 2t^2 - 1 + t^2 - 2t + 3 = 0
- t^2 - 2t + 3 = 0
Since (a + b + c = 0), use shortcut. The 2 real roots are: t = 1 and t = c/a = -3 (Rejected because < -1).

Within interval (-2pi, pi/2),

t = sin x = 1 --> x = -(3pi)/2 or x = pi/2

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