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Solve the equation $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = 8$ where $k$ is an integer and $x$ is a real number.

Attempt

We have that $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = \dfrac{\sin^{2k}(x)+\cos^{2k}(x)}{\cos^{2k}(x)\sin^{2k}(x)} = 8 \implies \sin^{2k}(x)+\cos^{2k}(x) - 8\cos^{2k}(x)\sin^{2k}(x) = 0.$ I get stuck here and am not sure how to proceed.

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    $\begingroup$ Hint: for $k>4$, you should find that the minimum of $\frac{1}{\sin^{2k}(x)}+\frac{1}{\cos^{2k}(x)}$ is greater than $8$. For $k\le 0$, the maximum is at most $2$. So you only have to consider $k=1,2,3,4$. $\endgroup$ – Joey Zou Dec 23 '15 at 1:50
  • $\begingroup$ How did you get the maximums and minimums? $\endgroup$ – John Ryan Dec 23 '15 at 1:51
  • $\begingroup$ Sorry, the minimum is greater than $8$ for $k>2$, so you actually only need to check $k=1$ and $k=2$. $\endgroup$ – Joey Zou Dec 23 '15 at 1:57
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    $\begingroup$ If you know calculus, you can just find the extrema by taking the derivative of the expression, although it may be a bit tedious that way. Alternatively, using the AM-GM inequality to get that $\frac{1}{\sin^{2k}(x)}+\frac{1}{\cos^{2k}(x)}\ge\frac{2}{|\sin^{k}(x)\cos^{k}(x)|} = \frac{2*2^{k}}{|2\sin{x}\cos{x}|^k} = \frac{2^{k+1}}{|\sin{2x}|}\ge 2^{k+1}$. So if $k>2$, then the expression is greater than $2^{2+1} = 8$. $\endgroup$ – Joey Zou Dec 23 '15 at 1:58
  • $\begingroup$ @JohnRyan Take the first and second derivatives. $\endgroup$ – John Dec 23 '15 at 2:01
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We have $$8=\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)} \geq \dfrac{2}{\left\vert\sin^k(x)\cos^k(x)\right\vert} = \dfrac{2^{k+1}}{\left\vert\sin^k(2x)\right\vert} \geq 2^{k+1}$$ Hence, we have $k+1 \leq 3 \implies k \leq 2$.

If $k=0$, we have $\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)}=2$.

If $k < 0$, we have $$\sin^{-2k}(x) + \cos^{-2k}(x) \leq \sin^2(x) + \cos^2(x) = 1$$ Hence, the only options are $k=1$ and $k=2$.

If $k=2$, we see that $$8 = \dfrac1{\sin^4(x)} + \dfrac1{\cos^4(x)} \geq \dfrac8{\sin^2(2x)} \implies \sin^2(2x) \geq 1 \implies \sin^2(2x) = 1$$

This implies that $$2x = n\pi + \dfrac{\pi}2 \implies x = \dfrac{n\pi}2 + \dfrac{\pi}4$$

When $k=1$, we have that $$\dfrac1{\sin^2(x)} + \dfrac1{\cos^2(x)} = \dfrac1{\sin^2(x)\cos^2(x)} = \dfrac4{\sin^2(2x)} = 8$$ Hence, $$\sin^2(2x) = \dfrac12 \implies 2x = n\pi \pm \dfrac{\pi}4 \implies x = \dfrac{n\pi}2 \pm \dfrac{\pi}8$$

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    $\begingroup$ how did you conclude if $k<0$ that $$\sin^{-2k}(x) + \cos^{-2k}(x) \leq \sin^2(x) + \cos^2(x) = 1?$$ $\endgroup$ – John Ryan Dec 23 '15 at 2:26
  • $\begingroup$ @JohnRyan $0\le \sin^2 x\le 1$, therefore $\left(\sin^2 x\right)^{-k}\le \sin^2 x$ for all $k\le -1$ (same for $\cos$), because $f(x)=a^x$ is non-increasing in $(0,+\infty)$ when $0\le a\le 1$. $\endgroup$ – user236182 Dec 23 '15 at 2:32
  • $\begingroup$ @user236182 Isn't $\dfrac{1}{0.5} > 0.5$ though, for example? $\endgroup$ – John Ryan Dec 23 '15 at 2:34
  • $\begingroup$ @JohnRyan It's unrelated. $-k\ge 1$, not $-k\le -1$. $\endgroup$ – user236182 Dec 23 '15 at 2:36
  • $\begingroup$ I believe Leg should've stated that the solution set is the union of solutions. $\endgroup$ – John Ryan Dec 23 '15 at 2:58

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