Question

how to solve 2cos^2 x- 3cosx +1=0 on the interval {0,2pi}

Answer 1

let `cos(x) = a`
then solve the quadratic....

Answer 2

I did it and,i got cos x=1 and cos x = 1/2 but I need to know on the interval {0,2pi} what is the solution

Answer 3

You are correct so far.

Answer 4

Now you have
\(\cos x = 1\) or \(\cos x = \dfrac{1}{2} \)

Answer 5

Let's deal with \(\cos x = 1\) first.
Here is a unit circle.
Which angle measure from 0 to \(2\pi\) has a cosine of 1?