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I solved like this:

\begin{align*}4^{\sin^2x}+4^{\cos^2x}=8&\Rightarrow4^{\sin^2x}+4^{1-\sin^2x}=8\\ &\Rightarrow4^{\sin^2x}+\frac{4}{4^{\sin^2x}}=8 |\cdot4^{\sin^2x}\\ &\Rightarrow4^{2\sin^2x}-8\cdot4^{\sin^2x}+4=0\\ y=4^{\sin^2x}&\Rightarrow y^2-8y+4=0\\ &\Rightarrow\Delta=64-16=48\\ &\Rightarrow y_{1,2}=\frac{8\pm 4\sqrt{3}}{2}\\ &\Rightarrow y_{1,2}=4\pm 2\sqrt{3}\\ &\Rightarrow 4^{\sin^2x}=4 \pm 2\sqrt{3} \end{align*}

But now I'm stuck.

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Note: $8\times 4^{\sin^2 x}\neq 8^{\sin^2x}$. You have $$8\times 4^{\sin^2 x} = 2^3\times 2^{2\sin^2x} = 2^{3+2\sin^2 x} = 2\times 4^{1+\sin^2 x}\neq 8^{\sin^2 x}$$ so your third line is already incorrect. E.g., if $x=\pi/2$, then $8\times 4^{\sin^2x} = 8\times 4 = 32$, but $8^{\sin^2x} = 8$. – Arturo Magidin Nov 22 '11 at 17:55
You should ask yourself whether you could have $4^a+4^b=8$ with $a<1$ and $b<1$. – Marco Disce Nov 22 '11 at 18:04
So take for example the case $4^{\sin^2 x}=4+2\sqrt{3}>4$. Then you should have $\sin^2 x >1$. Contradiction. If $4^{\sin^2 x}=4-2\sqrt{3}$ then $4^{\cos^2 x}=4+2\sqrt{3}>4^1$. Again contradiction – Beni Bogosel Nov 22 '11 at 18:04
It was a typographical error the third line. I corrected it. However, it doesn't change the rest of the steps because I considered it to be $8\cdot4^{sin^2x}$ – Daniel Nov 22 '11 at 18:21
@Daniel: Yes, I noticed that you seemed to make two errors that canceled each other; I'll edit my answer accordingly. – Arturo Magidin Nov 22 '11 at 18:23
up vote 5 down vote accepted

(Deleted comments refering to a previous version of the question which included a pair of seeming arithmetical errors). You have the equation: $$4^{\sin^2 x} = 4\pm 2\sqrt{3}.$$ Taking logarithms, you would get $$\log(4^{\sin^2 x}) = \log(4\pm 2\sqrt{3}),$$ or equivalently, $$\sin^2x \log(4) = \log(4\pm 2\sqrt{3}),$$ and hence $$\sin^2 x = \frac{\log (4\pm 2\sqrt{3})}{\log 4}.$$ However, $$\frac{\log (4+ 2\sqrt{3})}{\log 4}\approx 1.44998,\qquad\text{and}\qquad \frac{\log(4 - 2\sqrt{3})}{\log 4}\approx -0.44998,$$ so neither quantity can equal $\sin^2 x$ with $x$ real (since $0\leq \sin^2 x \leq 1$ for all real numbers $x$). So there are no (real) solutions.

Of course, Beni Bogossel's answer is better, but I write this in case you are interested in seeing how to carry your process all the way to the correct conclusion.

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The error in the third line was just a mistake. On paper I wrote correctly. – Daniel Nov 22 '11 at 18:22

This equality cannot happen. Since $\sin^2 x , \cos^2 x \in [0,1]$ it follows that $$4^{\sin^2 x}+4^{\cos^2 x}\leq 4+4=8$$ with equality when $\sin^2x=\cos^2x=1$. The last equality is impossible.

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it might still have complex roots – lurscher Nov 22 '11 at 18:03
@lurscher: I know, but i don't think that the OP wants complex roots. This would make the question a complex analysis one, not algebra. – Beni Bogosel Nov 22 '11 at 18:09
Nice illustration of "look before bringing out the machinery." – André Nicolas Nov 22 '11 at 18:11

If you do want complex roots, one of the possible values of $x$ is $\arcsin(\sqrt{\log_4(4+2\sqrt{3})}) = \pi/2 + i t$ where $$ t = \frac{1}{2} \ \ln \left( {\frac {\ln \left( 2 \right) }{\ln \left( 2+\sqrt {3} \right) +\sqrt {\ln \left( 4+2\,\sqrt {3} \right) \ln \left( 1+1/ 2\,\sqrt {3} \right) }}} \right) \approx -.6285882035 $$

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