solve for x: 2sin²(x) + 3cos(x) - 3 = 0

Answer 1

2sin²(x) + 3cos(x) - 3 = 0

Using the Pythagorean Identity sin²(x) + cos²(x) = 1, solve it for sin²(x) to get sin²(x) = 1 - cos²(x) then substitute "1 - cos²(x)" for "sin²(x)" as follows:

2(1 - cos²(x)) + 3cos(x) - 3 = 0

2 - 2cos²(x) + 3cos(x) - 3 = 0

-2cos²(x) + 3cos(x) - 1 = 0

2cos²(x) - 3cos(x) + 1 = 0

Let w = cos(x) so the above equation becomes:

2w² - 3w + 1 = 0 [now factor to get:

(2w - 1)(w - 1) = 0 [set each binomial equal to zero to get:

w = 1/2 and w = 1 [now back substitute to get:

cos(x) = 1/2 and cos(x) = 1

Using the unit circle x = π/3, x = 5π/3, and x = 0

These are the values for a domain of [0, 2π). If there are no domain restrictions then you would need to include all trips around the circle like this:

x = π/3 + 2πk, x = 5π/3 + 2πk, and x = 0 + 2πk where k is any integer

Answer 2

2sinx^2 + 3cosx - 3 = 0

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

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

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

2cosx^2 - 3cosx + 1 = 0

(2cosx - 1)(cosx - 1) = 0

2cosx - 1 = 0 or cosx - 1 = 0

cosx = 1/2 or cosx = 1

x = 0, 60, 180 and 300 degrees