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)=\frac23`

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

`\cos(2t)=\frac23\implies2t=\cos^(-1)\left(\frac23\right)+2n\pi`

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

`\implies t=\frac12\cos^(-1)\left(\frac23\right)+n\pi`

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

`t=\frac12\cos^(-1)\left(\frac23\right)\text{ and }t=\frac12\cos^(-1)\left(\frac23\right)+\pi`