Double-Angle Identities

Question

Solve  for all solutions 

Answer 1

This really isn't a double angle identity problem.

Just solve for 2x, then divide your answer by 2 to get x

6(1-sin²2x)^1/2 = 6sin2x + 2

divide by 6, then square both sides

1-sin²2x = sin²2x + 2sin2x/3 + 1/9

2sin²2x + (2/3)sin2x -8/9 = 0

use the quadratic formula

2x = 31.4 degrees or 58.6+180 = 238.6 degrees

divide by 2

x = 15.7 degrees and 119.3 degrees. Convert to radians if you want

only solutions are in Quadrant I and III for 2x

the only solutions for x are in Quadrant I & II.

You never use a double or half angle trig formula

The solution for 2x is very close to a 30-60-90 triangle in Quadrant I and III

Let A=2x

CosA = SinA + 1/3

sqr3/2 = 1/2 + 1/3

.866 = almost .833

Cos2x has to be a little smaller, and if you make 30 degrees a little larger, such as 31.4 the cosine is a little smaller

Answer 2

6 cos(2x) = 6 sin²(x) + 2

6 (1 - 2sin²x) = 6 sin²x + 2

18 sin²x = 4

sin²x = 2/9

sin x = (2/9)^0.5

Then, you can solve the equation by finding what is x.