8cos^2 theta - 4 sin^2 theta = 6?

Question

solve for the angle theta in the following equation

Answer 1

8cos^2 theta - 4 sin^2 theta = 6
8(1-sin^2 theta)-4sin^2 theta =6
8-8*sin^2 theta-4sin^2 theta =6
-12sin^2 theta=-2
sin^2 theta = 1/6
sin theta= +/- sqrt(1/6)
theta 1=0.167448079 rad =9.594068227 degrees
theta 2=-0.167448079 rad=-9.594068227 degrees

Answer 2

Add 8cos^2 Theta to both sides, then the equastion will be simplify to:

8 - 4sin^2 theta = 6 + 8sin^2 theta

8cos^2 theta + 8 sin^2 theta= 8(cos^2 theta + sin^2 theta) = 8

Answer 3

We need to convert either cos^2 theta to sin or sin^2 theta to cos.

Let's convert sin^2 theta

According to the identity:
sin^2 theta + cos^2 theta = 1

Therefore:
sin^2 theta = 1 - cos^2 theta

Substitute it in the main equation:
Thefore:
8cos^2 theta - 4*( 1 - cos^2 theta) = 6

Let cos^2 theta = x
So:

8x - 4*(1 - x) = 6
8x - 4 + 4x = 6
12x = 10
x = 5/6

Hence:
cos^2 theta = 5/6
cos theta = +-sqrt(5/6)
cos theta = sqrt(5/6) or cos theta = -sqrt(5/6)
theta = 24.095 deg or theta = 155.905 deg

This is the answer.

Answer 4

let's use x instead of Θ since it's easier to type:

8 cos² x - 4 sin² x = 6
8 cos² x - 4(1 - cos² x) - 6 = 0
8 cos² x - 4 + 4 cos² x - 6 = 0
12 cos² x - 10 = 0
cos² x = 5/6
cos x = ±√30 / 6 = ±0.91287
x = 0.4205 radians (reference angle) or
x = 2.7211 radians (quadrant 2) or
x = 3.5621 radians (quadrant 3) or
x = 5.8627 radians (quadrant 4)

Answer 5

8 cos² Θ - 4 sin² Θ = 6

Divide both sides by 2
4 cos² Θ - 2 sin² Θ = 3

Substitute cos² Θ = 1 - sin² Θ
4(1 - sin² Θ) - 2 sin² Θ = 3

Distribute
4 - 4 sin² Θ - 2 sin² Θ = 3

Combine terms and transpose
-6 sin² Θ = -1

Divide both sides by -6
sin² Θ = 1/6

Get the square root of both sides
sin Θ = ± √6/6

Therefore,
Θ = arcsin √6/6 or Θ = arcsin - √6/6

^_^

Answer 6

Sove for θ.

8cos²θ - 4sin²θ = 6
4cos²θ - 2sin²θ = 3
(2cos²θ - 1) + 2(cos²θ - sin²θ) = 2
cos 2θ + 2cos 2θ = 2
3cos 2θ = 2
cos 2θ = 2/3
2θ = arccos(2/3)
θ = (1/2)arccos(2/3) ≈ 24.094843°