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
 A: By trial and error we can easily see that $x=-\pi/2$ and $x=0$ are solutions.  We shall prove that there are no more solutions, except of course for adding $2n\pi$.
First, if $-\pi/2<x<0$ then 
$$-1<\sin x\quad\hbox{and}\quad\cos x<1\ .$$
Multiplying these inequalities by $\sin^2x$ and $\cos^2x$, each of which is positive, gives
$$-\sin^2x<\sin^3x\quad\hbox{and}\quad\cos^3x<\cos^2x\ .$$
Therefore we have
$$\cos^3 x<\cos^2x=1-\sin^2x<1+\sin^3x\ ,$$
and it is not possible that $\cos^3x-\sin^3x=1$.
That was the hard part!  For $0<x\le\pi$ we have
$$\cos^3x<1\le1+\sin^3x$$
and for $\pi<x<3\pi/2$ we have
$$\cos^3x<0<1+\sin^3x\ .$$
So the only solutions are $x=2n\pi$ and $x=2n\pi-\pi/2$.
A: HINT:
$$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)=\frac{\sqrt2\cos\left(\frac\pi4+x\right)(2+\sin2x)}2$$
Set $\displaystyle\frac\pi4+x=y$
$\displaystyle\implies\sin2x=\sin2\left(y-\frac\pi4\right)=-\sin\left(\frac\pi2-2y\right)=-\cos2y=-(2\cos^2y-1)$
On rearrangement on the given relation, there will be a cubic equation in $\cos y$ 
By observation,  one the values of $\cos y$ is $\dfrac1{\sqrt2}$
The rest is to find the resultant Quadratic Equation 
A: Hint:
$$\cos^3(x)-\sin^3(x) = (\cos x  - \sin x )(\cos^2  x + \cos x\sin x + \sin^2 x)$$
So we have:
$$(\cos x  - \sin x )(\color{Blue}{\cos^2  x} + \cos x\sin x + \color{Blue}{\sin^2 x}) = 1$$
$$(\cos x  - \sin x )(\color{Blue}{1} + \cos x\sin x) = 1$$
A: Let $\cos x = t$.  Then you have $t^3-1 = (1-t^2)^{3/2}$.  Squaring and simplifying, we have
$$(1-t^3)^2 = (1-t^2)^3 \iff (1-t)^2(1+t+t^2)^2=(1-t)^3(1+t)^3 $$
So $t=1$ is a solution, or $(1+t+t^2)^2 = (1-t)(1+t)^3 \iff t^2(2t^2+4t+3) = 0$.  
The last equation yields $t=0$ as the last quadratic is $2(t+1)^2+1 > 0$.
Thus we have $\cos x = 0, 1$ as the only possible real solutions, and correspondingly $\sin x = -1, 0$...
A: $$\cos^3x-\sin^3x=(\cos x-\sin x)(1+\sin x\cos x)$$
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
