How many solutions of $x$ are available of the equation $3\tan x+x^3 =2 $ in range of $[0,\pi/4]$? $3\tan(x)+x^3 =2$ in range of $[0,\frac{\pi}{4}$]? I mean how to start with this. I'm not given the calculator to put distinct values. So what's the key idea to solve it? Should I put $\tan(x) = x$ for smaller values of $x$? Or anything else? 
 A: Let $f(x)=3\tan(x)+x^3$.

Since $\tan(x)$ and $x^3$ are both strictly increasing on $[0,{\large{\frac{\pi}{4}}}]$, it follows that $f$ is strictly increasing on $[0,{\large{\frac{\pi}{4}}}]$

Since $f$ is strictly increasing on $[0,{\large{\frac{\pi}{4}}}]$, it follows that the equation $f(x) = 2$ has at most one solution for $x\in [0,{\large{\frac{\pi}{4}}}]$.

Noting that $f(0)=0$, and
$$f({\small{\frac{\pi}{4}}})=3\tan({\small{\frac{\pi}{4}}})+({\small{\frac{\pi}{4}}})^3 = 3+({\small{\frac{\pi}{4}}})^3 > 2$$
it follows, by the Intermediate Value Theorem, that the equation $f(x) = 2$ has at least one solution for $x\in [0,{\large{\frac{\pi}{4}}}]$.

Therefore, the equation $f(x)=2$ has exactly one solution, for $x\in [0,{\large{\frac{\pi}{4}}}]$.
A: Only one.
$$f(x)=3\tan x+x^3$$ 
...is monotonously increasing function in range [0, $\pi/4$].
$$f(0)<2, \quad f(\pi/4)>2$$
So there can be only one value of $x$ for which $f(x)=2$.
The following plot just confirms that:

A: Hint
Consider that you are looking for the zero of function
$$f(x)=3 \tan (x)+x^3-2$$ for which $$f'(x)=3 \sec ^2(x)+3 x^2$$ I am sure that you can  conclude.
