How do you solve for x in 2sin^(2)(2x)+3cos(2x)=3

Question

I have tried solving with half-angle formula and factoring and can not get an answer

Answer 1

2sin²(2x) + 3cos(2x) = 3

 

2(1 - cos²(2x)) + 3cos(2x) = 3

 

-2cos²(2x) + 3cos(2x) +2 = 3

 

-2cos²(2x + 3cos(2x) -1 = 0

 

2cos²(2x) - 3cos(2x) + 1 = 0

 

(cos(2x) -1)(2cos(2x) - 1) = 0

 

cos(2x) = 1 or cos(2x) = 1/2

 

If cos(2x) = 1, then 2x = 2kπ, k = 0, ±1, ±2, ...

 

                              So, x = kπ

 

If cos(2x) = 1/2, then 2x = π/3 + 2kπ or 5π/3 + 2kπ

 

                              So, x = π/6 + kπ or 5π/6 + kπ

 

Solutions between 0 and 2π:  0, π, 2π, π/6, 7π/6, 5π/6, 11π/6 

 

Answer 2

Using the basic Pythagorean Identity, replace the squared sine by 1 - equivalent squared cosine, and you will soon get this equation reduced to cos(2x) = 1 which you will readily solve.