$ (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 $

See https://math.stackexchange.com/questions/1330422/how-to-solve-this-equation-for-x-left-sqrt2-sqrt3-rightx-left-sq or https://math.stackexchange.com/questions/1237837/solve-left-sqrt34-sqrt15-rightx-left-sqrt34-sqrt15-right?noredirect=1&lq=1 — lab bhattacharjee, Oct 13 '18 at 17:11

Dr. Sonnhard Graubner · Accepted Answer · 2018-10-13T16:34:58.050

2

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

edited Oct 13 '18 at 16:34

answered Oct 13 '18 at 16:17

is that first = supposed to be a + ? – SomeGuyDoingMaths Oct 13 '18 at 16:29
It is an equal sign, multiply it out, $$25-24=1$$ – Dr. Sonnhard Graubner Oct 13 '18 at 16:32
no sorry in the second equation is that supposed to be a "=" after the first part – SomeGuyDoingMaths Oct 13 '18 at 16:34
No it must be a plus sign, a typo sorry – Dr. Sonnhard Graubner Oct 13 '18 at 16:35
Then I just finish it like with the quadratic formula im assuming – SomeGuyDoingMaths Oct 13 '18 at 16:36
Yes it will be a quadratic equation – Dr. Sonnhard Graubner Oct 13 '18 at 16:36

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