4 sin 2x + 3 = 4 cos x + 6 sin xHow do i solve this trigonometry question? I can not seem to find a to replace identities that would be able to...

One way to solve this problem is to use these
identities:

sin 2x = 2 sin x*cos
x

sin x = 2 tan(x/2)/[1+(tan
x/2)^2]

cos x = [1-(tan x/2)^2]/[1+(tan
x/2)^2]

We'll replace tan x/2 by
t:

sin x = 2 t/(1+t^2)

cos x =
(1-t^2)/(1+t^2)

We'll replace these identities into
equation:

16 t(1-t^2)/(1+t^2)^2 + 3 = 4(1-t^2)/(1+t^2) +
12t/(1+t^2)

16 t*(1-t^2) + 3(1+t^2)^2 = 4(1-t^4) +
12t*(1+t^2)

We'll remove the brackets and we'll combine
like terms:

7t^4 - 28t^3 + 6t^2 + 4t - 1 =
0

Therefore, to solve the trigonometric
equation, you'll have to determine the zeroes of the polynomial 7t^4 - 28t^3 + 6t^2 + 4t
- 1 = 0.