Question

How do you solve `2cosx-sqrt3=0`?

Answer 1

`pi/6+-n2pi`, `(5pi)/6+-n2pi`

Explanation:

We have:

`2cosx-sqrt3=0`

We can simplify it this way:

`2cosx=sqrt3`

`cosx=sqrt3/2`

`x=pi/6=30^o`

Ok, so we have our reference angle. Now let's apply it to a coordinate system.

Cosine is positive in Q1 and in Q4.

In Q1, we have the angle already as `pi/6`. Since there is no constraint on the solution, the general solution is `pi/6+-n2pi`, where n is a natural number

In Q4, the angle is `2pi-pi/6=(5pi)/6` and again we need to show the general solution of `(5pi)/6+-n2pi` with n again being a natural number.