Cubic trig equation I'm trying to solve the following trig equation:
$\cos^3(x)-\sin^3(x)=1$
I set up the substitutions $a=\cos(x)$ and $b=\sin(x)$ and, playing with trig identities, got as far as $a^3+a^2b-b-1=0$, but not sure how to continue. Is there a way to factor this so I can use the zero product property to solve?
Thanks for any help/guidance!
P

$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)$$
If we set $\displaystyle\cos x-\sin x=u, u^2=1-2\sin x\cos x\implies (\cos x-\sin x)(1+\sin x\cos x)=u\left(1+\frac{1-u^2}2\right)=\frac{3u-u^3}2$
So, the problem reduces to  $\displaystyle u^3-3u+2=0$
Clearly, $1$ is a root and $\displaystyle\displaystyle\frac{u^3-3u+2}{u-1}=u^2+u-2=(u+2)(u-1)$
$\displaystyle\implies u^3-3u+2=(u+2)(u-1)^2$
If $\displaystyle u+2=0, u=-2,\cos x-\sin x=-2,$   $\displaystyle\cos x-\sin x=\sqrt2\cos\left(x+\frac\pi4\right)$
$\displaystyle\implies-\sqrt2\le\cos x-\sin x\le\sqrt2$
So, $u=\cos x-\sin x$ must be $\displaystyle=1\implies \cos x-\sin x=1\iff \sqrt2\cos\left(x+\frac\pi4\right)=1$
$\displaystyle\implies\cos\left(x+\frac\pi4\right)=\frac1{\sqrt2}=\cos\frac\pi4$
$\displaystyle\implies x+\frac\pi4=2m\pi\pm \frac\pi4$ where $m$ is any integer
