# How do you solve #Cos(2x)cos(x)-sin(2x)sin(x)=0#?

##### 1 Answer

May 27, 2018

We have:

#cos(2x)cosx = sin(2x)sinx#

#1 = tan(2x)tanx#

Now we have to make some manipulations using

#1 = (2sinxcosx)/(cos^2x- sin^2x) * sinx/cosx#

#cos^2x -sin^2x = 2sin^2x#

#1 - sin^2x - sin^2x = 2sin^2x#

#1 = 4sin^2x#

#1/4 = sin^2x#

#sinx = +- 1/2#

It's now clear that

Hopefully this helps!