Solving Trig Function with both $\sin$ and $\cos$ I am trying to understand the function $h(t) = 65 + 36\sin\left(\frac{3t}{2}\right) - 15\cos\left(\frac{3t}{2}\right)$, where $t$ is the time in seconds and $\frac{3t}{2}$ is expressed in radians.
A question asks that I find the times in the first revolution when the height is exactly $65$m. Using basic algebra I have been able to solve in order to get time which is $t = 0.26319$. But the question asks for times in the first revolution and I am having trouble understanding how to find these values.
Hence, I need to find the time to complete one revolution and the angular speed of the function. 
How would you determine the times when the height is $65$m and from that be able to determine the time it takes to complete one revolution and the angular speed of the function?
I assume that this equation is not in the form of $a + b\sin(cx)$?
 A: Hint
use $${a\over \sqrt{a^2+b^2}}\cos u+{b\over \sqrt{a^2+b^2}}\sin u=\cos(u-\theta)$$where$$\theta =\cos^{-1}{a\over \sqrt{a^2+b^2}}=\sin^{-1}{b\over \sqrt{a^2+b^2}}$$
A: You are looking for $t$ such that
$$
\begin{align}
&\quad h(t) = 65\\
\Rightarrow &\quad 36\sin\left(\frac{3t}{2}\right) = 15\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12\sin\left(\frac{3t}{2}\right) = 5\cos\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 12^2\sin^2\left(\frac{3t}{2}\right) = 5^2\cos^2\left(\frac{3t}{2}\right)\\
\Rightarrow &\quad 144\sin^2\left(\frac{3t}{2}\right) = 25\left({1-\sin^2\left(\frac{3t}{2}\right)}\right)\\
\Rightarrow &\quad 169\sin^2\left(\frac{3t}{2}\right) = 25\\
\Rightarrow &\quad 13\sin\left(\frac{3t}{2}\right) = 5\\
\Rightarrow &\quad \sin\left(\frac{3t}{2}\right) = \frac{5}{13}\\
\Rightarrow &\quad \frac{3t}{2} = \sin^{-1}\left(\frac{5}{13}\right)\\
\Rightarrow &\quad t = \frac23\sin^{-1}\left(\frac{5}{13}\right)
\end{align}
$$
A: The equation has the classical form 
$$a\cos t+b\sin t=c.$$
Among million other ways, you can solve with
$$(a+b\tan t)^2=\frac{c^2}{\cos^2t}=c^2\tan^2t+c^2$$
or
$$(c^2-b^2)\tan^2t-2ab\tan t+c^2-a^2=0,$$
giving
$$\tan t=\frac{ab\pm c\sqrt{a^2+b^2-c^2}}{c^2-b^2}.$$
