Question

How do you solve `4cos^2x-1=0`?

Answer 1

`x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3`

Explanation:

This is a trivial quadratic equation in `cosx`

`4cos^2x-1=0`
`:. 4cos^2x = 1`
`:. cos^2x = 1/4`
`:. cosx = +-1/2`

In order to solve this you need to be familiar with the graphs of the trig functions along with common values:

The range of soutions is not specified in the question, so lets assume we want a solutio in radians, and `-2pi <= x <= 2pi`

Consider, `cos=1/2 => x=-2pi+pi/3,-pi/3,pi/3, 2pi-pi/3`
`:. x=(-5pi)/3,(-pi)/3,pi/3, (5pi)/3`

Consider, `cos=-1/2 => x=-2pi+(2pi)/3,(-2pi)/3,(2pi)/3, 2pi-(2pi)/3`
`:. x=(-4pi)/3,(-2pi)/3,(2pi)/3, (4pi)/3`

So the solutions are:
`:. x=+-(4pi)/3,+-(2pi)/3,+-(5pi)/3,+-(pi)/3`