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

1 Answer
Noah G
May 27, 2018

We have:

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

#1 = tan(2x)tanx#

Now we have to make some manipulations using #tan(2x) = (2sinxcosx)/(cos^2x- sin^2x)#

#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 #x = pi/6, (5pi)/6, (7pi)/6 and (11pi)/6#. In general form, these are #pi/6 + pin# and #(5pi)/6 + pin#. However we must also include when #cosx = 0#, namely when #x = pi/2 + pin#. We can confirm graphically.

enter image source here

Hopefully this helps!

Related questions
See all questions in Solving Trigonometric Equations
Impact of this question
6770 views around the world
You can reuse this answer
Creative Commons License