TITLE: Show that $f(x)=\sqrt 2$ has no solutions $f(x)=\sin x\cos x(2+\sin x)$ and $x\in [0,\frac{\pi}{2}]$.
QUESTION [15 upvotes]: Show that $f(x)=\sqrt 2$ has no solutions when
$$f(x)=\sin x\cos x(2+\sin x)$$
and $x\in [0,\frac{\pi}{2}]$.

My attempt:
Since $f(x)$ is continuous in its domain, it is enough to show that maximum value attained by $f$ in $[0,\frac{\pi}{2}]$ is less than $\sqrt 2$.
Also, since $f(0)=f(\pi/2)=0$, it must hold true that $f'(c)=0$ for some $c\in[0,\frac{\pi}{2}]$ (Rolle's Theorem). And since $f(\pi/6)>0$, there has to exist a maxima.
But differentiation doesn't help me here, because when I set $f'(x)$ to $0$, I get (on rearrangement and using basic trigonometry) a cubic in $\sin x$:
$$3t^3+4t^2-2t-2=0$$
where $t=\sin x$.
The above cubic doesn't have any rational roots and I'm stuck.
Any help will be great. Thanks!
Edit:
I created this question by myself to solve in a pen-paper test. So methods that involve usage of calculators are useless. No offence.
P.S. Please keep in mind that I'm barely seventeen, so no highly advance math please!

REPLY [7 votes]: We have
$$f(x)=2\sin(x)\cos(x)+\sin^2(x)\cos(x)=\sin(2x)+\cos(x)-\cos^3(x).$$
Now for $0\leq x\leq\pi/2$ we have $0\leq\sin(2x)\leq1$ and the function $x\mapsto\cos(x)-\cos^3(x)$ has its maximum at $\arctan(\sqrt2)$, its value is
$$\frac{2}{3\sqrt3}=\frac{2\sqrt3}9<\frac{2\cdot1.8}9=0.4.$$  From here
$$\sin(2x)+\cos(x)-\cos^3(x)\leq\sqrt2.$$