The course of the function $f(x) = \sin^3 (x) + \cos^3 (x)$. First, sorry for my english, hope you understand.
so I'm examining the course of this function $f(x) = \sin^3 (x) + \cos^3 (x)$ where I need to find everything I can about this function (e.g. definition field, range of values, inflection points, asymptots, global and local max/min, etc.). Now I'm struggling with 3 tasks.

*

*Prove there are no limits in $+\infty$ by showing there are different limits of $f(\pi/2+2k\pi)$ and $f(2k\pi)$ where $k$ goes to $+\infty$. Is this right?
 Are the results of the limits different?


*How to get inflection points? I know I get them by solving the 2nd derivative of $f(x)$ which I think I have but I don't know how to get the exact values.
graph and inflex points 


*Find intersection of $f(x)$ with axes. I got to the point where I have two equations and I need to prove why is it like that. 
Thank you in advance
edit: Thank you all for helping me with this problem. I've used @Robert Lee solution in the end.
 A: COMMENT.-(1)Any periodic function defined over $\mathbb R$ has a limit when $x\to\infty$$\space (f$ non-constant).
(2) $f(x)=(\sin(x)+\cos(x))(1-\sin(x)\cos(x))$ so the first factor gives the zeros of the function $x=-\dfrac{\pi}{4}+k\pi$ where $k\in\mathbb Z$. (the second factor clearly has
non-real roots).
(3) Inflection points: you have $f''(x)=-3(\cos(x)-\sin(x))^2(\cos(x)+\sin(x))=0$ find yourself $f'''(x)$ and determine this thirds question
A: (1) the idea is right, note that $\sin(k\pi) = 0$ and $\cos(k\pi) = (-1)^k$
(2) Note
$$
\begin{split}
f'(x) &= 3\sin^2 x \cos x - 3\cos^2x\sin x 
       = 3 \sin x \cos x (\sin x - \cos x)\\
f''(x) &= \ldots = -3 (\cos x - \sin x)^2(\cos x + \sin x)\\
\end{split}
$$
and now solving $f''(x)=0$ is trivial.
(3) your argument makes sense, if you are asking something else, please clarify it
A: 1)
Notice that if $k \in \mathbb{Z}$ we have that
$$
\sin\left( 2k\pi - \frac{\pi}{2} \right) = -\cos\left(2k\pi \right)= -1
$$
and
$$
\cos\left( 2k\pi - \frac{\pi}{2} \right) = \sin\left(2k\pi \right)= 0
$$
for the first limit. For the second, notice that
$$
\sin\left( 2k\pi  \right) = 0
$$
and
$$
\cos\left( 2k\pi  \right) =1
$$
Finally, notice that
$$
\lim_{k \to \infty} (-1)^3 + 0^3 = -1 \color{red}{\neq} 1 = \lim_{k \to \infty} (0)^3 + 1^3
$$

2)
This one's a bit tricky. By repeated application of the chain rule and then using trigonometric identities we get the following:
\begin{align*}
\frac{d^2}{dx^2}\sin^3(x) + \cos^3(x) &= 3 \left[2 \sin(x) \cos^2(x) - \underbrace{\sin^3(x)}_{\color{blue}{\left(1-\cos^2(x)\right) \sin(x)}} + 2 \cos(x) \sin^2(x) - \underbrace{\cos^3(x)}_{\color{blue}{\left(1-\sin^2(x)\right) \cos(x)}}\right] \\
& =  3 \left[2 \sin(x) \cos^2(x) \color{blue}{-\sin(x) + \cos^2(x) \sin(x)} + 2 \cos(x) \sin^2(x) \color{blue}{-\cos(x) + \sin^2(x) \cos(x)}\right]\\
& = 3 \left[\color{blue}{3} \sin(x) \cos^2(x)  + \color{blue}{3}  \cos(x) \sin^2(x) - \sin(x) - \cos(x)\right]\\
& =\frac{3}{2} \left[6 \sin(x) \cos^2(x)  + 6  \cos(x) \sin^2(x) -2 \sin(x) - 2\cos(x)\right]\\
& =\frac{3}{\sqrt{2}\sqrt{2}}\left( \sin(x) + \cos(x)\right)\left(6 \  \underbrace{\sin(x) \cos(x)}_{\color{purple}{\frac{\sin(2x)}{2}}} -2 \right)\\
& =\frac{3}{\sqrt{2}}\left( \sin(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\cos\left(\frac{\pi}{4}\right)}} + \cos(x)\underbrace{\frac{1}{\sqrt{2}}}_{\color{green}{\sin\left(\frac{\pi}{4}\right)}}\right)\left(\color{purple}{3}\sin(2x) -2 \right)\\
& = \frac{3}{\sqrt{2}}\sin\left(x + \frac{\pi}{4}\right)\left(3\sin(2x) -2 \right)
\end{align*}
We can set this last equation equal to $0$, which would them imply that each of the factors is equal to $0$ separately. Recalling that
$$
\sin(\theta) = \alpha \color{blue}{\iff} \theta = (-1)^n\arcsin\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z}  \tag{1}
$$
We then see that the inflection points are at
\begin{align}
\sin\left(x + \frac{\pi}{4}\right) = 0 &\iff x + \frac{\pi}{4} = (-1)^n\arcsin\left(0\right) + \pi n\\
&\iff x + \frac{\pi}{4} = \pi n\\
&\iff \boxed{x  = \pi n -  \frac{\pi}{4}}, \quad n \in \mathbb{Z} 
\end{align}
And also at
\begin{align}
3\sin(2x) -2 = 0 & \iff \sin(2x) = \frac{2}{3}\\
&\iff 2x = (-1)^n\arcsin\left(\frac{2}{3}\right) + \pi n\\
&\iff x = \boxed{\frac{(-1)^n}{2}\arcsin\left(\frac{2}{3}\right) + \frac{\pi}{2} n},\quad n \in \mathbb{Z} 
\end{align}

3)
Your solution is correct. The zeros of the function are at the points where
$$
\sin(x) = - \cos(x) \color{blue}{\implies}\tan(x) = -1
$$
And recalling that
$$
\tan(\theta) = \alpha \color{blue}{\iff} \theta = \arctan\left(\alpha\right) + \pi n, \quad n \in \mathbb{Z}  \tag{2}
$$
We see that the zeros are at the points
\begin{align}
\tan(x) = -1 & \iff  x = \arctan(-1) + \pi n\\
& \iff  x = \boxed{-\frac{\pi}{4} + \pi n}, \quad n \in \mathbb{Z}
\end{align}
Which is exactly the same set that you propose in your solution since, if we start counting the integers from one over by setting
$$
n = k+1 \in \mathbb{Z}
$$
we get
\begin{align}
x &=-\frac{\pi}{4} + \pi n\\
&=-\frac{\pi}{4} + \pi (k+1)\\
&=-\frac{\pi}{4} + \pi k+ \pi\\
&=\pi\left(1-\frac{1}{4}\right) + \pi k\\
&=\frac{3\pi}{4} + \pi k, \qquad k \in \mathbb{Z}
\end{align}
Also, if you want to argue why $1 = \sin(x)\cos(x)$ has no solutions, recall the previously used identity that $\sin(x)\cos(x) = \frac{\sin(2x)}{2}$. So you want to show that
$$
\frac{\sin(2x)}{2} = 1 \color{blue}{\implies} \sin(2x) = 2
$$
but recalling that the sine function can only output values in the interval $[-1,1]$, since $2 \notin [-1,1] $ the above equation has no solutions.

If you have any questions about the solutions please let me know!
