How do you find all #sin6x+sin2x=0# in the interval #[0,2pi)#?

1 Answer
Feb 10, 2017

#0, pi/4, pi/2, (3pi)/4, 2pi#

Explanation:

Use trig identity:
#sin a + sin b = 2sin ((a +b)/2)cos ((a - b)/2)#
In this case -->
f(x) = sin 6x + sin 2x = 2sin (4x).cos (2x) = 0

a. sin 4x = 0
Unit circle gives 3 solutions:
4x = 0 --> x = 0
#4x = pi# --> #x = pi/4#
#4x = 2pi# --># x = (2pi)/4 = pi/2#
b. cos 2x = 0
trig unit circle gives 2 solutions:
#2x = pi/2# --> #x = pi/4#
#2x = (3pi)/2# --> #x = (3pi)/4#

All answers for #(0, 2pi)#
#0, pi/4, pi/2, (3pi)/4, 2pi#