MHB Trigonometry Challenge: Can You Solve This Equation?

[anemone]

Solve the equation $\sin^7 x+\dfrac{1}{\sin^3 x}=\cos^7 x+\dfrac{1}{\cos^3 x}$.

 

[Opalg]

Solve the equation $\sin^7 x+\dfrac{1}{\sin^3 x}=\cos^7 x+\dfrac{1}{\cos^3 x}$.

[sp]Multiply out the fractions: $$\sin^{10}\!x\cos^3\!x + \cos^3\!x = \cos^{10}\!x\sin^3\!x + \sin^3\!x,$$ $$\sin^{10}\!x\cos^3\!x - \cos^{10}\!x\sin^3\!x = \sin^3\!x - \cos^3\!x,$$ $$\sin^3\!x \cos^3\!x(\sin^7\!x - \cos^7\!x) = \sin^3x - \cos^3\!x,$$ $$\begin{aligned}\sin^3\!x &\cos^3\!x(\sin x - \cos x)(\sin^6\!x + \sin^5\!x\cos x + \sin^4\!x\cos^2\!x + \sin^3\!x\cos^3\!x + \sin^2\!x\cos^4\!x + \sin x\cos^5\!x + \cos^6\!x)\\ &= (\sin x - \cos x)(\sin^2\!x + \sin x\cos x + \cos^2\!x). \end{aligned}$$ One possibility (which in fact will turn out to be the only possibility) is $\sin x = \cos x$, or $x = \bigl(n+\frac14\bigr)\pi$. Otherwise, divide both sides by $\sin x - \cos x$: $$\sin^3\!x \cos^3\!x \bigl((\sin^4\!x + \cos^4\!x)(\sin^2\!x + \sin x\cos x + \cos^2\!x) + \sin^3\!x \cos^3\!x\bigr) = \sin^2\!x + \sin x\cos x + \cos^2\!x.$$ Now let $z = \sin x\cos x$. Notice that $z = \frac12\sin2x$, so that $|z| \leqslant \frac12.$ Also, $\sin^4\!x + \cos^4\!x = (\sin^2\!x+\cos^2\!x)^2 - 2\sin^2\!x\cos^2\!x = 1 - 2z^2$; and $ \sin^2\!x + \sin x\cos x + \cos^2\!x = 1+z.$ So the equation becomes $$z^3\bigl((1-2z^2)(1+z) + z^3\bigr) = 1+z,$$ $$z^6 - 2z^5 + z^4 + z^3 - z = 1.$$ But $|z^6 - 2z^5 + z^4 + z^3 - z| \leqslant |z|^6 + 2|z|^5 + |z|^4 + |z|^3 + |z| \leqslant \frac1{64} + \frac1{16} + \frac1{16} + \frac18 + \frac12 <1.$ So there are no solutions for $z$, showing that there are no further solutions for $x$.[/sp]

 

[anemone]

[sp]Multiply out the fractions: $$\sin^{10}\!x\cos^3\!x + \cos^3\!x = \cos^{10}\!x\sin^3\!x + \sin^3\!x,$$ $$\sin^{10}\!x\cos^3\!x - \cos^{10}\!x\sin^3\!x = \sin^3\!x - \cos^3\!x,$$ $$\sin^3\!x \cos^3\!x(\sin^7\!x - \cos^7\!x) = \sin^3x - \cos^3\!x,$$ $$\begin{aligned}\sin^3\!x &\cos^3\!x(\sin x - \cos x)(\sin^6\!x + \sin^5\!x\cos x + \sin^4\!x\cos^2\!x + \sin^3\!x\cos^3\!x + \sin^2\!x\cos^4\!x + \sin x\cos^5\!x + \cos^6\!x)\\ &= (\sin x - \cos x)(\sin^2\!x + \sin x\cos x + \cos^2\!x). \end{aligned}$$ One possibility (which in fact will turn out to be the only possibility) is $\sin x = \cos x$, or $x = \bigl(n+\frac14\bigr)\pi$. Otherwise, divide both sides by $\sin x - \cos x$: $$\sin^3\!x \cos^3\!x \bigl((\sin^4\!x + \cos^4\!x)(\sin^2\!x + \sin x\cos x + \cos^2\!x) + \sin^3\!x \cos^3\!x\bigr) = \sin^2\!x + \sin x\cos x + \cos^2\!x.$$ Now let $z = \sin x\cos x$. Notice that $z = \frac12\sin2x$, so that $|z| \leqslant \frac12.$ Also, $\sin^4\!x + \cos^4\!x = (\sin^2\!x+\cos^2\!x)^2 - 2\sin^2\!x\cos^2\!x = 1 - 2z^2$; and $ \sin^2\!x + \sin x\cos x + \cos^2\!x = 1+z.$ So the equation becomes $$z^3\bigl((1-2z^2)(1+z) + z^3\bigr) = 1+z,$$ $$z^6 - 2z^5 + z^4 + z^3 - z = 1.$$ But $|z^6 - 2z^5 + z^4 + z^3 - z| \leqslant |z|^6 + 2|z|^5 + |z|^4 + |z|^3 + |z| \leqslant \frac1{64} + \frac1{16} + \frac1{16} + \frac18 + \frac12 <1.$ So there are no solutions for $z$, showing that there are no further solutions for $x$.[/sp]

Neat and elegant solution, especially on the second part how you ruled out the possibility that there will be any solution(s) from the second situation! Thank you Opalg!