Trigonometric Equations/ Quadratic Formula/ Not exact values?

Question

1) 8sin^2(x)- 4cos(x)-3=0

2) 8cos^2(x)+4cos(x)-5=0

 

How do i solve.

Round answers to nearest tenth o degree

Answer 1

1) 8sin^2(x)- 4cos(x)-3=0

8sin²(x) - 4cos(x) - 3 = 0

Note: sin²A + cos²A = 1 than sin²A = 1 - cos²A

8(1 - cos²x) - 4cos(x) - 3 = 0

Simplify

8 - 8cos²x - 4cos(x) - 3 = 0

- 8cos²x - 4cos(x) - 3 + 8 = 0

- 8cos²x - 4cos(x) + 5 = 0

Multiply each side by negative one.

8cos²x + 4cos(x) - 5 = 0

Let  cos(x) = t

Than

8t² + 4t + 5 = 0

Quadratic formula

Conspire the equation ax²+bx+c = 0

So substitute a = 8, b = 4, and c = 4 above the formula

But t = cos(x)

There fore