$ (5+2\sqrt6)^{\sin x} +(5-2\sqrt6)^{\sin x} = 2\sqrt3 $ , where $ 0 ≤ x≤ 360 $

Question

There is something I haven't picked up on, a hint would be appreciated

Given that $(\sqrt3+\sqrt2)^2 = (5+2\sqrt6)$ and $ (\sqrt3-\sqrt2)^2 = (5-2\sqrt6)$

Find the values of x for which$ (5+2\sqrt6)^{\sin x} +(5-2\sqrt6)^{\sin x} = 2\sqrt3 $ , where $ 0 0 ≤ x≤ 360 $

Answer 1

Hint: Use that $$(5+2\sqrt{6})(5-2\sqrt{6})=1$$ Then you will get $$(5+2\sqrt{6})^{\sin(x)}+\frac{1}{(5+2\sqrt{6})^{\sin(x)}}=2\sqrt{3}$$