Question

How do you solve `2+ cos(2x)=3 cos (x)` over the interval 0 to 2pi?

Answer 1

`x=0, pi/3`

Explanation:

We want a function in terms of either `cos(x)` or `cos( 2x)`, but not both. We can use the double angle formula to convert `cos(2x)` into an expression with `cos(x)`.

Double Angle Formula
`cos(2x) = 2cos^2(x) -1`

Applying the double angle formula to our function gives;

`2+2cos^2(x) -1 = 3cos(x)`

Or, after a little rearranging;

`2cos^2(x) - 3cos(x) +1 = 0`

We can use the quadratic formula to factor this expression.

Quadratic Formula
`(-b+-sqrt(b^2-4ac))/(2a)` where `ax^2 + bx +c = 0`

Plugging in values for our function;

`cos(x)=(3+-sqrt((-3)^2-4(2)(1)))/(2(2))`

`cos(x)=(3+-sqrt(9-8))/4`

`cos(x)=(3+-sqrt(1))/4`

`cos(x)=1, 1/2`

A quick glance at a unit circle will show that;

`cos(0)=1`
`cos(pi/3)=1/2`

So;

`x=0, pi/3`