# Solving for angle given equation

#### Ilikebugs

how to solve for x in the interval [-pi,pi] for the equation 3sin(2x)+sqrt(3)cos(2x)=-3? I tried using identities but I can't get rid of one of the sin or cos.

#### Plato

MHF Helper
how to solve for x in the interval [-pi,pi] for the equation 3sin(2x)+sqrt(3)cos(2x)=-3? I tried using identities but I can't get rid of one of the sin or cos.
Look at this solution.

topsquark

#### Ilikebugs

I need to pay to get the step by step solution

#### Cervesa

rearrange
$3(\sin(2x) + 1) = -\sqrt{3} \cos(2x)$
note that $\sin(2x)+1 \ge 0 \implies \cos(2x) \le 0$

square
$9(\sin^2(2x)+2\sin(2x)+1) = 3\cos^2(2x) = 3[1 - \sin^2(2x)]$
$3(\sin^2(2x)+2\sin(2x)+1) = \cos^2(2x) = 1 - \sin^2(2x)$
$4\sin^2(2x) + 6\sin(2x)+2 = 0$
$2\sin^2(2x)+3\sin(2x)+1 = 0$
$[2\sin(2x)+1][\sin(2x)+1] = 0$

$\sin(2x) = -\dfrac{1}{2} \text{ and } \sin(2x) = -1$
note $-\pi < x < \pi \implies -2\pi < 2x < 2\pi$

take it from here?

topsquark

#### Ilikebugs

Does squaring give extraneous solutions?

#### Cervesa

Does squaring give extraneous solutions?
yes, that's the reason for my first "note" stating $\cos(2x) \le 0$