I saw this problem on an instagram feed
$81^{\sin^{2}(x)} + 81^{\cos^{2}(x)} = 30$$$81^{\sin^{2}(x)} + 81^{\cos^{2}(x)} = 30$$
solve for x
And I saw that most people had solutions where you plug in x = 30 degrees or $\frac{\pi}{6}$ and the like. I get that if you assume that the two numbers added are integers, you can sort of back-calculate and say that
$81^{\frac{1}{4}} = 3 $
and
$81^{\frac{3}{4}} = 27$
so that gets you an answer that works.
But I was trying to come up with a good algebraic way to solve it -- I thought of using logs, but I don't think you can really do that here. I did see on method that says
let $u = 81^{\cos^{2}(x)}$ and you'd get, since $\cos^{2}(x) = 1 - \sin^{2}(x)$
$81^{1-\cos^2{x}} = \frac{81}{81^{\cos^2{x}}}$
So we can substitute in
$\frac{81}{u} + u = 30$
$81 + u^2 = 30u$
$u^2 - 30u + 81 = 0$
Solve this by factoring $(u - 3)(u - 27) = 0$ so solutions are at 3 and 27
that means $ u = 81^{cos^{2}(x)} = 3 $ therefore $\cos(x) = \frac{\sqrt{3}}{2}$
And it all works. But now that I seem to have figured this out I'd be curious as to why it works; or rather, is there some other method that would make sense? Again I thought of logs or something, but I feel like this method was "cheating" somehow...