Question

How do you solve `Cos(2x)cos(x)-sin(2x)sin(x)=0` over the interval 0 to 2pi?

Answer 1

Solutions are:
`pi/6`, `pi/2`, `(5pi)/6`, `(7pi)/6`, `(3pi)/2`, `(11pi)/6`

Explanation:

This is just an exercise on a simple trigonometric formula of a cosine of a sum of two angles.

As is known,
`cos(a+b)=cos(a)*cos(b)-sin(a)*sin(b)`

Assuming `a=2x` and `b=x`, we get:
`cos(2x+x)=cos(2x)*cos(x)-sin(2x)*sin(x)`

Therefore, on the left of the equation we have `cos(2x+x)=cos(3x)` and the given equation is equivalent to
`cos(3x)=0`

Cosine of an angle (in radians) is equal to `0` at angles measured `pi/2+pi *n` for all integer `n` (positive, 0 and negative).
Therefore, the common solution of our equation is
`3x=pi/2+pi*n`
or
`x=pi/6+pi/3*n`
Varying integer `n` from 0 forward, we get the following solutions in the interval `[0,2pi]`:
`pi/6`, `pi/2`, `(5pi)/6`, `(7pi)/6`, `(3pi)/2`, `(11pi)/6`