Solving exponential (of trigonometric functions) equation

[anemone]

Hi MHB,

Solve $(2+ \sqrt{2})^{(\sin x)^2}-(2- \sqrt{2})^{(\cos x)^2}=\left( 1+ \dfrac{1}{\sqrt{2}} \right)^{\cos 2x} -(2-\sqrt{2})^{\cos 2x}$.

This problem vexes me much because the only way that I could think of to solve this problem would be by substituting $(\sin x)^2=u$, and from there, I gotten another ugly equation $\left( 2^u-(2-\sqrt{2})=\dfrac{(2-\sqrt{2})^u(\sqrt{2}+1)^{1-2u}}{\sqrt{2}^{1-2u}}-\dfrac{2-\sqrt{2}}{(2-\sqrt{2})^u} \right)$, of which I don't think I am heading in the right direction...

I normally would employ another method to approach the problem since the first trial failed me, but the thing is, other than trying to solve this problem by the substitution skill, I don't think I can come up with another idea of how to crack it. Hence, I brought it up here with the hope to get some insights from the forum, and if any of you could provide me with any ideas of how to even start to work with this problem, I would greatly appreciate it.:)

 

[mente oscura]

Hi MHB,

Solve $(2+ \sqrt{2})^{(\sin x)^2}-(2- \sqrt{2})^{(\cos x)^2}=\left( 1+ \dfrac{1}{\sqrt{2}} \right)^{\cos 2x} -(2-\sqrt{2})^{\cos 2x}$.

Hello.

I don't know if you will serve:

[tex]a=2+\sqrt{2}[/tex]

[tex]1+\dfrac{1}{\sqrt{2}}=\dfrac{a}{2}[/tex]

[tex]2-\sqrt{2}=\dfrac{2}{a}[/tex]

[tex]\cos^2x=b[/tex]

[tex]\sin^2x=1-b[/tex]

[tex]\cos(2x)=2b-1[/tex]

Regards.