# How do you solve sin 4x + sin 2x = 0 using the product and sum formulas?

Nov 27, 2014

$\setminus R i g h t a r r o w \setminus \sin 2 \left(2 x\right) + \setminus \sin 2 x = 0$
$\setminus R i g h t a r r o w 2 \setminus \sin 2 x \setminus \cos 2 x + \setminus \sin 2 x = 0$
$\setminus R i g h t a r r o w \setminus \sin 2 x \left(2 \setminus \cos 2 x + 1\right) = 0$

Case 1: $\setminus \sin 2 x = 0$,
$\setminus R i g h t a r r o w 2 x = 0 \setminus R i g h t a r r o w x = 0 , 2 \setminus \pi , 4 \setminus \pi , \ldots$

Case 2: $\left(2 \setminus \cos 2 x + 1\right) = 0$
$\setminus R i g h t a r r o w 2 \setminus \cos 2 x = - 1 \setminus R i g h t a r r o w \setminus \cos 2 x = - \setminus \frac{1}{2}$
$\setminus R i g h t a r r o w 2 x = \setminus \frac{2 \setminus \pi}{3} , \setminus \frac{4 \setminus \pi}{3} , \setminus \frac{8 \setminus \pi}{3} , \setminus \frac{10 \setminus \pi}{3} , \ldots$
$\setminus R i g h t a r r o w x = \setminus \frac{\setminus \pi}{3} , \setminus \frac{2 \setminus \pi}{3} , \setminus \frac{4 \setminus \pi}{3} , \setminus \frac{5 \setminus \pi}{3} , \ldots$

Cosine is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd quadrants, hence the choice of angles.