Question

How do I solve the equation in an interval from 0 to 2π ? 9 cos 2t = 6​

Answer 1

$9\cos(2t)=6\implies\cos(2t)=\dfrac23$

Using the fact that cos is 2π-periodic, we have

$\cos(2t)=\dfrac23\implies2t=\cos^{-1}\left(\dfrac23\right)+2n\pi$

That is, $\cos(\theta+2n\pi)=\cos\theta$ for any $\theta$ and integer $n$.

$\implies t=\dfrac12\cos^{-1}\left(\dfrac23\right)+n\pi$

We get 2 solutions in the interval [0, 2π] for $n=0$ and $n=1$,

$t=\dfrac12\cos^{-1}\left(\dfrac23\right)\text{ and }t=\dfrac12\cos^{-1}\left(\dfrac23\right)+\pi$