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

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