Trig Question, Solve For X

Question

Solve for x where pi/2 ≤ x ≤ 3pi/2.

- 4cos^2 x + 4sqrt(2) sin x + 6 = 0.

When I did the problem I got 5pi/6, but I am not sure if that is correct.

Answer 1

Remembering that sin²(x) + cos²(x) = 1 gives us cos²(x) = 1 - sin²(x)

Substituting "1 - sin²(x)" into the equation in place of cos²(x) gives:

- 4[1 - sin²(x)] + 4√2sin(x) + 6 = 0

Multiplying this out, combining like terms and simplifying, we get:

-4 + 4sin²(x) + 4√2sin(x) + 6 = 0

4sin²(x) + 4√2sin(x) + 2 = 0

2sin²(x) + 2√2sin(x) + 1 = 0

Let w = sin(x), then it becomes:

2w² + 2√2w + 1 = 0

Now, using the quadratic formula to solve:

w = (-2√2 ± √[(2√2)² - 4(2)(1)])/(2*2)

w = (-2√2 ± √(8-8))/4

w = -2√2/4

w = -√2/2

Since w = sin(x) then sin(x) = -√2/2

That occurs when x = 5π/4