Solving this problem algebraically: $81^{\sin^{2}(x)} + 81^{\cos^{2}(x)} = 30 $

Question

I saw this problem on an instagram feed

$$81^{\sin^{2}(x)} + 81^{\cos^{2}(x)} = 30$$

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...

Answer 1

This is a very fine method, indeed we have that by $u=81^{\cos^{2}(x)}\neq 0$ and using $\cos^2 x+\sin^2 x=1$

$$81^{\sin^{2}(x)} + 81^{\cos^{2}(x)} = 30 \iff \frac{81}u+u=30$$

that is the two equations are completely equivalent and since

$$\frac{81}u+u=30 \iff u=3 \;\lor\; u=27$$

we obtain

• $u=81^{\cos^{2}(x)}=3 \implies \cos^2 x=\frac14$
• $u=81^{\cos^{2}(x)}=27 \implies \cos^2 x=\frac34$

which are indeed the two solutions.