I got stuck on this equation. Could you help me a bit and tell me how should I approach it? The answer of it is 3/4 and -3/4.
$$4^{\sin^2(\pi x)}+4^{\cos^2(\pi x)}=-8x^2+12|x|-{1\over2}$$
Solving fancy equation
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transcendental-equations
1 Answers
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Let $S=4^{\sin^2\pi x}$. Since $\cos^2\pi x=1-\sin^2\pi x$, the left hand side can be rewritten as $S+4S^{-1}$. The smallest (positive) value that $S+4S^{-1}$ takes on occurs when $S=2$, with value $2+4/2=4$.
As for the right hand side, completing the square leads to
$$-8x^2+12|x|-{1\over2}=4-{1\over2}(4|x|-3)^2$$
which means that $4$ is the largest value the right hand side takes on, and it only occurs when $4|x|-3=0$, which is to say, when $x=\pm3/4$. It remains to check that $S=2$ for these values of $x$, which it does: $\sin^2(\pm3\pi/4)=(1/\sqrt2)^2=1/2$, so $S=4^{\sin^2(\pm3\pi/4)}=4^{1/2}=\sqrt4=2$.
The key fact, that $S+4S^{-1}$ is minimized at $S=2$, can be established in various ways. The easiest might be to differentiate the function $f(S)=S+4S^{-1}$: $$f'(S)=1-4S^{-2}=0\implies S=2$$