Solve for X Problem

Question

Solve for x where pi/2 ≤ x ≤ 3pi/2.

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

Enter your answer as a fraction.

Answer 1

-4cos²x + 4√2sinx +6 =0

1. Use the Pythagorean Identity sin²x + cos²x = 1: -4(1-sin²x) + 4√2sinx + 6 =0
2. Distribute -4: -4 + 4sin²x + 4√2sinx + 6 =0
3. combine like terms: 4sin²x + 4√2sinx + 2 =0
4. Factor out the GCF, which is 2: 2(2sin²x +2√2sinx +1) =0
5. Use the Quadratic Formula. a=2, b=2√2, c=1: sinx = [-2√2 ± √((2√2)²-4(2)(1))]/(2(2))
6. (2√2)² = 4(2) = 8. So the discriminant, the part under the square root, simplifies to 8-8, which equals 0. So the fraction is now simplified to: sinx = (-2√2)/4, which simplifies to sinx = -√2/2.
7. Determine when sinx is equal to -√2/2. Sine is equal to ±√2/2 at all angles that have a reference angle of π/4. Sine is negative in quadrants 3 and 4. However, the original domain was π/2 ≤ x ≤ 3π/2, which is in quadrants 2 and 3. So to satisfy the original equation and domain restrictions, x must equal 5π/4.

Answer 2

I tried to simplify and possibly factor this but couldn't.

So I graphed it and x has values of (2.16 ≤ x ≤ 2.49) rounded to two decimals.