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

1 Answer
Nov 27, 2014

#\Rightarrow\sin 2(2x) + \sin 2x=0#
#\Rightarrow 2\sin 2x\cos 2x + \sin 2x = 0#
#\Rightarrow \sin 2x(2\cos 2x + 1)=0#

Case 1: #\sin 2x = 0#,
#\Rightarrow 2x = 0\Rightarrow x=0, 2\pi, 4\pi,...#

Case 2: #(2\cos 2x + 1) = 0#
#\Rightarrow 2\cos 2x = -1 \Rightarrow \cos 2x = -\frac{1}{2}#
#\Rightarrow 2x=\frac{2\pi}{3}, \frac{4\pi}{3}, \frac{8\pi}{3},\frac{10\pi}{3},...#
#\Rightarrow x=\frac{\pi}{3}, \frac{2\pi}{3}, \frac{4\pi}{3},\frac{5\pi}{3},...#

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