Is my solution for finding the values of t where h is equal to 1.5 correct?

[Bushy]

I have $h = 1+0.6 \cos \left( \frac{\pi t}{2}\right)$ and need to find where $h=1.5$ for $t\in [0,4]$

The period of the function is 4 and I get solutions of

$\displaystyle 1.5 = 1+0.6 \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle0.5 = 0.6 \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle \frac{5}{6} = \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle \cos^{-1}\frac{5}{6} = \frac{\pi t}{2}$

$\displaystyle t = \frac{2\times \cos^{-1}\frac{5}{6}}{\pi} \approx 0.373, 4-0.373 $

Can anyone find a mistake ere?

 

[Sudharaka]

I have $h = 1+0.6 \cos \left( \frac{\pi t}{2}\right)$ and need to find where $h=1.5$ for $t\in [0,4]$

The period of the function is 4 and I get solutions of

$\displaystyle 1.5 = 1+0.6 \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle0.5 = 0.6 \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle \frac{5}{6} = \cos \left( \frac{\pi t}{2}\right)$

$\displaystyle \cos^{-1}\frac{5}{6} = \frac{\pi t}{2}$

$\displaystyle t = \frac{2\times \cos^{-1}\frac{5}{6}}{\pi} \approx 0.373, 4-0.373 $

Can anyone find a mistake ere?

Hi Bushy, :)

\[\frac{5}{6} = \cos \left( \frac{\pi t}{2}\right)\]

\[\Rightarrow\frac{\pi t}{2}=2n\pi\pm\cos^{-1}\left( \frac{5}{6}\right)\]

\[\Rightarrow t=4n\pm\frac{2\cos^{-1}\left( \frac{5}{6}\right)}{\pi}\mbox{ where }n\in\mathbb{Z}\]

Since \(t\in[0,4]\) the only solutions are,

\[t=3.627141002645325\mbox{ and }t=0.37285899735467\]

So your solutions are correct. :)

Kind Regards,
Sudharaka.