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!