How do you solve #Sin(3x) - sin(6x) = 0#?

1 Answer
May 14, 2016

Answer:

#0, pi/3, (2pi)/3, pi/9, and (5pi)/9#

Explanation:

Use the trig identity: sin 2a + 2sin a.cos a.
Replace in the equation sin (6x) by 2sin (3x).cos (3x) -->
sin (3x) - 2sin (3x).cos (3x) = 0
sin (3x)(1 - 2cos 3x) = 0
a. sin 3x = 0 --> 3x = 0 and #3x = pi#, and #3x = 2pi#, that give as answers-->
#x = 0; x = pi/3, and x = (2pi)/3#
b. #cos (3x) = 1/2# --> #3x = +- pi/3# -->
#x = (pi/3)/3 = pi/9# and #x = ((5pi)/3)/3 = (5pi)/9#
Note: the arc #-(pi/3)# is co-terminal to the arc #(5pi)/3.#