# Thread: VERY HARD Trigonometric Equation!!

1. ## VERY HARD Trigonometric Equation!!

$2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3=0$

Solve for x.

Maple gave 60, 120, 150 & -150 degrees as the answers.

I tried to find a solution but in vain.

2. ## Re: VERY HARD Trigonometric Equation!!

1. Use an identity to deal with sin(2x).
2. Use as much algebra as you can.

3. ## Re: VERY HARD Trigonometric Equation!!

Originally Posted by phys251
1. Use an identity to deal with sin(2x).
2. Use as much algebra as you can.
1. I did already.
2. " ... as much Algebra as you can." = infinity ...

4. ## Re: VERY HARD Trigonometric Equation!!

After using an identity for sin(2x) bring all the terms with $2\sqrt{3}$ to one side of the equals sign and then square both sides. The identity $sin^2x+cos^2x=1$ will be helpful later.

5. ## Re: VERY HARD Trigonometric Equation!!

haha. THanks. Got it ... I didn't tried that one bcoz I proceeded to more complex solution quickly.

6. ## Re: VERY HARD Trigonometric Equation!!

Hello, Kaloda!

$\text{Solve for }x\!:\;2\sin{2x}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0$

We have: . $2\underbrace{\sin{2x}}+2\sqrt{3}\sin{x}-2\sqrt{3}\cos{x}-3\:=\:0$
Then: . $2\overbrace{(2\sin x\cos x)} + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0$

. . . . . . $4\sin x\cos x + 2\sqrt{3}\sin x - 2\sqrt{3}\cos x - 3 \:=\:0$

Factor: . $2\sin x\left(2\cos x + \sqrt{3}\right) - \sqrt{3}\left(2\cos x + \sqrt{3}\right) \:=\:0$

Factor: . $\left(2\cos x + \sqrt{3}\right)\left(2\sin x - \sqrt{3}\right) \:=\:0$

Therefore:

. . $2\cos x + \sqrt{3} \:=\:0 \quad\Rightarrow\quad \cos x \:=\:-\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:150^o,\,210^o$

. . $2\sin x - \sqrt{3} \:=\:0 \quad\Rightarrow\quad \sin x \:=\:\frac{\sqrt{3}}{2} \quad\Rightarrow\quad x \:=\:60^o,\,120^o$