# 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 solve the question.

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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.**