How do you solve #sin 2theta sin theta = cos theta#?

1 Answer
May 16, 2015

From the formula for #sin (2theta)# we have

#sin 2theta = 2sin theta cos theta#

From Pythagoras theorem we have

#sin^2 theta + cos^2 theta = 1#

So

#sin 2theta sin theta#
#= (2sin theta cos theta)sin theta#
#= 2sin^2 theta cos theta#
#=2(1-cos^2 theta)cos theta#

Putting this together with your equation, we get

#2(1-cos^2 theta)cos theta = cos theta#

If #cos theta = 0# then both sides will be zero.
So some solutions to the original problem are:

#theta = pi/2 + npi# for all #n# in #ZZ#.

On the other hand, if #cos theta != 0#, divide both sides of the equation by #cos theta# to get

#2(1-cos^2 theta) = 1#

Divide both sides by 2 to get

#1-cos^2 theta = 1/2#

So #cos^2 theta = 1/2# and #cos theta = +-1/sqrt(2)#

This is true for

#theta = pi/4 + (npi)/2# for all #n# in #ZZ#.