Question

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

Answer 1

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.

Hopefully this helps!