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$4^{\sin^2x}+4^{\cos^2x}=8$

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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    @Daniel: Yes, I noticed that you seemed to make two errors that canceled each other; I'll edit my answer accordingly.2011-11-22

3 Answers 3

4

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 $

13

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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    Nice illustration of "look before bringing out the machinery."2011-11-22
5

(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.2011-11-22