Solve $\sin^{3}x+\cos^{3}x=1$ 
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$

$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
 A: Both of $\sin(x),\cos(x)$ must be nonnegative since, if one of them was negative, the equation $\sin^3(x)+\cos^3(x)=1$ would imply that the other one is more than $1$, contradiction.

Thus, we have $0\le \sin(x) \le 1$ and $0\le \cos(x)\le 1$.

If both of $\sin(x),\cos(x)$ are less than $1$, then, since they are both nonnegative, we would have 
$$
\begin{cases}
0\le \sin^3(x) < \sin^2(x)\\[4pt]
0\le \cos^3(x) < \cos^2(x)\\
\end{cases}
$$
but that would imply
$$\sin^3(x)+\cos^3(x) < \sin^2(x)+\cos^2(x)=1$$
contradiction.

It follows that one of $\sin(x),\cos(x)$ must be equal to $1$, and the other must be equal to zero.

From that information, I'm sure you can finish the solution.
A: Hint:
Let $\sin x+\cos x=t\implies t^2=?$
$$1=\dfrac{t\{2-(t^2-1)\}}2\iff t^3-3t+2=0$$
Clearly, $t=1$ a solution
A: Let use by $t= \tan (x/2)$


*

*$\sin x=\frac{2t}{1+t^2}$

*$\cos x=\frac{1-t^2}{1+t^2}$
to obtain 
$$2t^6-8t^3+6t^2=0$$
$$\iff t^2(t^4-4t+3)=t^2(t-1)^2(t^2+2t+3)=0$$
and since $t^2+2t+3>0$ the solutions are


*

*$t=0 \implies \frac x 2=k\pi\implies x=2k\pi$

*$t=1 \implies \frac x 2=\frac{\pi}4+k\pi \implies x=\frac{\pi}2+2k\pi $
A: As an alternative by
$$\sin^3 x+ \cos ^3 x=1 \iff \sin x\cdot \sin^2+\cos x\cdot \cos^2 x=1$$
since $\sin^2 x+ \cos ^2 x=1$, the given equality is a weighted mean of $\sin x$ and $\cos x$ which holds if and only if 


*

*$\sin x=1,\,\cos x=0$
or


*

*$\cos x=1,\,\sin x=0$
A: Hint $$\sin^3(x)+\cos^3(x)=\sin^2(x)+\cos^2(x).$$
