Question

Solve the following equations for all solutions of x

2sin^2x+3cos-3=0

Answer 1

Recall that sin^2x + cos^2x = 1 so sin^2x = 1 - cos^2x

2sin^2 x + 3cosx - 3 = 0

2(1 - cos^2x) + 3cosx - 3 = 0

2cos^2x - 3cosx + 1 = 0

This is just a quadratic equation in cosx so let u = cosx

2u^2 - 3u + 1 = 0

This factors as

(2u-1)(u-1) = 0

So 2u = 1 and u = 1 or 2cosx = 1 and cosx = 1

cosx = 1 when x = 0
cosx = 1/2 when x = pi/3 and 5pi/3

So x = {0, pi/3, 5pi/3} is the solution set for your equation.

Answer 2

Thank you so much.