Question

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

Answer 1

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`