Find the solution of this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$ 
Possible Duplicate:
Solve $(\sqrt{5+2\sqrt{6}})^{x}+(\sqrt{5-2\sqrt{6}})^{x}=10$. 

Help me solve this equation $(\sqrt{2-\sqrt 3})^x+(\sqrt{2+\sqrt 3})^x=4$

Let $t_1=(\sqrt{2+\sqrt3})^x$, and $t_2=(\sqrt{2-\sqrt3})^x$. Now the given equation is:
$t_1+t_2=4$..........(1)
Above it follows that:
$t_1\cdot t_2=(\sqrt{2+\sqrt 3})^x\cdot(\sqrt{2-\sqrt3})^x=(\sqrt{({2+\sqrt3})({2-\sqrt3}})^x=(\sqrt{2^2-(\sqrt3)^2})^x=(\sqrt{4-\sqrt 3^2})^x=(\sqrt{4-3})^x=1^x=1$ $\Rightarrow$ $t_1\cdot t_2=1$...........................(2)
For (1) and (2) we have:
$t_1+t_2=4$
$t_1\cdot t_2=1$
$\Rightarrow$ $t^2-4t+1=0$
For this quadratic equation have:
$t_{1,2}=\frac{4\pm\sqrt{16-4}}{2}$
$t_{1,2}=\frac{4\pm 2\sqrt{3}}{2}$
$t_1=2+\sqrt 3$, $t_2=2-\sqrt 3$
Now return the inital substition:
$t_1=(\sqrt{2+\sqrt3})^x$
$2+\sqrt 3=(\sqrt{2+\sqrt3})^x$ 
$(\sqrt{2+\sqrt 3})^2=(\sqrt{2+\sqrt3})^x$ $\Rightarrow$ $x=2$, and 
$(\sqrt{2+\sqrt3})^x=2-\sqrt 3$
$(\sqrt{2+\sqrt3})^x=\sqrt{(2+\sqrt 3)}^{-2}$ $\Rightarrow$ $x=-2$
Definitly $x=2$, and $x=-2$ is solve.
