Inversion of Trigonometric Equations I've been playing around with finding the domain-restricted inverses of trigonometric equations using the inverse trigonometric equations. One of the easier formulas I came up with was the formula for the inverse
$$a\cos^2x+b\sin^2x$$
The process I used to invert this was to use the pythagorean identities to turn it into a single trigonometric function:
$$=a(\cos^2x+\sin^2x)+(b-a)\sin^2x$$
$$=a(1)+(b-a)\sin^2x$$
$$=a+(b-a)\sin^2x$$
and so now I can easily say that the partial inverse is
$$\arcsin\sqrt{\frac{x-a}{b-a}}$$
The other form that I managed to find an inversion formula for was the form
$$a\cos x+b\sin x$$
My strategy for this one was to use the sum-angle identity after manipulating the expression a bit:
$$=\sqrt{a^2+b^2}\bigg(\frac{a}{\sqrt{a^2+b^2}}\cos x+\frac{b}{\sqrt{a^2+b^2}}\sin x\bigg)$$
$$=\sqrt{a^2+b^2}\bigg(\cos x\sin\arcsin\frac{a}{\sqrt{a^2+b^2}}+\sin x\cos\arccos\frac{b}{\sqrt{a^2+b^2}}\bigg)$$
$$=\sqrt{a^2+b^2}\bigg(\cos x\sin\arcsin\frac{a}{\sqrt{a^2+b^2}}+\sin x\cos\arcsin\frac{a}{\sqrt{a^2+b^2}}\bigg)$$
and now I use the sum-angle formula to reduce this to
$$=\sqrt{a^2+b^2}\sin\bigg(x+\arcsin\frac{a}{\sqrt{a^2+b^2}}\bigg)$$
and now we can easily find that the inverse is
$$\arcsin\bigg(\frac{x}{\sqrt{a^2+b^2}}\bigg)-\arcsin\bigg(\frac{a}{\sqrt{a^2+b^2}}\bigg)$$
However, the last one I've been working with is giving me a little bit of trouble. I can't figure out how to invert
$$a\cos x+b\sin^2 x$$
and based on the shapes of its graphs, I suspect some forms can't even be partially inverted this way. 
Am I on a wild goose chase? If not, does anybody have any hints?
One more question - does anybody know of other interesting expressions like my two examples that can be inverted? I really enjoy the puzzle of manipulating these expressions to get an inverse, but I don't want to waste my time on any impossible ones.
Thanks!
 A: Invert the equation $y=a \cos x +b \sin^2 x$ (for $x$)? 
Substitute $\sin^2 x = 1- \cos^2 x$, multiply by $-4b$ and complete the square.
\begin{eqnarray*}
\underbrace{4 b^2 \cos^2 x -4 ab \cos x + a^2}_{(2b \cos x-a)^2} =a^2 +4b^2 -4by \\
 x = \cos^{-1} \left( \frac{a \pm \sqrt{a^2+4 b^2 -4 b y}}{2b} \right)
\end{eqnarray*}
So in keeping with the question (where $x$ and $y$ are swapped ) the formula that you seek is 
\begin{eqnarray*}
 \color{red}{\cos^{-1} \left( \frac{a \pm \sqrt{a^2+4 b^2 -4 b x}}{2b} \right)}.
\end{eqnarray*}
FURTHER EXAMPLES : Some other nice examples of similar functions that can be inverted are ...
\begin{eqnarray*}
a \cos x +b \sin 2 x \,\,\,\,\, & \longleftrightarrow & \,\,\,\,\, ? 
\end{eqnarray*}
\begin{eqnarray*}
a \tan^2 x +b \sec^2 x \,\,\,\,\, & \longleftrightarrow & \,\,\,\,\, ? 
\end{eqnarray*}
\begin{eqnarray*}
a \tan^2 x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? 
\end{eqnarray*}
\begin{eqnarray*}
a \tan x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? \tiny{\text{HINT : convert to $ \cos$ and $\sin$ and then use the $cos$ half} angle formula.}
\end{eqnarray*}
\begin{eqnarray*}
a \cos(hx+j) +b \cos (hx+k) \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? 
\end{eqnarray*}
And here are the two from the question:
\begin{eqnarray*}
a \cos x +b \sin x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? 
\end{eqnarray*}
\begin{eqnarray*}
a \cos^2 x +b \sin^2 x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? 
\end{eqnarray*}
And another .
\begin{eqnarray*}
a \cos x +b \sec x \,\,\,\,\,\, & \longleftrightarrow & \,\,\,\,\,\, ? 
\end{eqnarray*}
