Show that $f(x)=\sqrt 2$ has no solutions $f(x)=\sin x\cos x(2+\sin x)$ and $x\in [0,\frac{\pi}{2}]$. 
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!
 A: Here's a solution based on the remark of Angina Seng. It's quite annoying and as such might not be useful to you, but I found it interesting.
We see that it suffices to show that the polynomial $f(s)=2-s^2(1-s^2)(2+s)^2$ is a sum of squares. We seek a representation as the sum of two squares of integer polynomials.
Note that $f(-2)=f(-1)=f(0)=f(1)=2$, so if $f(s)=P(s)^2+Q(s)^2$ we require that $P(s),Q(s)\in\{-1,1\}$ for each $s\in\{-2,-1,0,1\}$. Since $f$ is a degree $6$ monic polynomial, we must have that one of $\{P,Q\}$ is degree $3$ (say that's $P$) and one is lower degree (say it's $Q$). Since we can find values of $s$ for which $f(s)-1<0$, we can't set $Q$ to be constant at $1$ or $-1$, so $Q$ must be linear or quadratic. However, since $Q(-2),Q(-1),Q(0),Q(1)\in\{-1,1\}$, some value must occur twice as a value of $Q$, so $Q$ can't be linear and must be quadratic. Since it can't switch direction too many times, its values must be $1,-1,-1,1$ at $-2,-1,0,1$ (or the negatives of these), and so we want
$$Q(s)=s^2+s-1.$$
Now, we see that
$$P(s)^2=s^6+4s^5+2s^4-6s^3-3s^2+2s+1.$$
By looking at the $s^6$, $s^5$, $s$, and $1$ terms, we should have that the polynomial should start with $s^3+2s^2$ and end with $\pm(s+1)$. Ending with $+s+1$ is bad because then $P$ would have all positive coefficients, so we should have $P(s)=s^3+2s^2-s-1$, which does actually square to the above. Thus, we have
$$f(s)=(s^3+2s^2-s-1)^2+(s^2+s-1)^2.$$
Now all we need to do is show that $f(s)$ is never actually $0$. This is not too bad; assume that $s^2+s=1$ and $s^3+2s^2=s+1$. Then
$$s+1=s^3+2s^2=(s+2)(s^2)=(s+2)(1-s)=-s^2-s+2=1\implies s=0,$$
a contradiction.
A: 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
$$\sin(2x)+\cos(x)-\cos^3(x)\leq\sqrt2.$$
A: If you want to avoid numerical methods, note that the problem is equivalent to proving
$\sin^2x\cos^2x(2+\sin x)^2<2$ on $[0,\pi/2]$. This is the same as proving $s^2(1-s^2)(2+s)^2<2$ on $[0,1]$ where we set $s=\sin s$. You can prove this using rational
arithmetic by considering the Sturm sequence of the polynomial
$f(s)=2-s^2(1-s^2)(2+s)^2$, but I wouldn't like to do this by hand.
Plotting the graph of $f$ persuades me that $f(s)$ is positive for all real $s$.
You may be able to prove this by writing $f(s)$ as a positive linear combination
of squares of polynomials.
A: Consider the function $g(x)=sin(x)cos(x)$ independently. It can easily seen with calculus that the maximum value it attains is $=1/2$ (at $x=\pi/4+2n\pi$).
So, it immediately follows that $f(x)\leq \frac{1}{2}(2+sinx)=p(x)$, with the equality holding when $x=\pi/4+2n\pi$. Thus it is sufficient to show that $p(x)$ itself is less than $\sqrt2$ in the given interval at the $x$ for which $f(x)$ was maximum (that is, at the roots of the cubic you formed). So, we have -
$$\frac{1}{2}(2+t)<\sqrt2$$
$$t<2(\sqrt2-1)$$
Now, we have transformed the question into checking if the root of the cubic in the interval $[0,1]$ is $<2(\sqrt2-1)$.
Let the cubic be $T(t)$. By differentiating once and calculating the roots of the quadratic by the quadratic formula we find that one of its minima/maxima is necessarily $-ve$. Also, we notice that $T(0)$ is $-ve$ and $T(1)$ is $+ve$ . Thus, there are either $1$ or $3$ roots between $0$ and $1$. But, as we said right before one of the minima/maxima of the cubic is $-ve$, and for all $3$ roots to be in $[0,1]$ both the minima/maxima need to be in it as well. Thus, the cubic only has one root for $x\in[0,\frac{\pi}{2}]$.
Now, if the root was $<2(\sqrt2-1)$, $T(2(\sqrt2-1))$ would be $+ve$. This is fairly easy to compute by hand (especially since you only need to know the sign), and what you'll see is that it indeed is true- $T(2(\sqrt2-1))>0$.
A: $f(x) = sin(2x) + \frac{sin(2x)sin(x)}{2} = sin(2x)(1 + \frac{sin(x)}{2})$
Since $x \in [0,\frac{\pi}{2}]$, $sin(x)$ and $sin(2x)$ are both positive.
So, we can apply, AM-GM inequality to say that :
$\sqrt(f(x)) <= \frac{(sin(2x) + \frac{sin(x)}{2} + 1)}{2}$
So, now we try to find the max value of $g(x) = sin(2x) + \frac{sin(x)}{2} + 1$
$g'(x) = 2cos(2x) + \frac{cos(x)}{2} = 2(2cos^2(x) - 1) + \frac{cos(x)}{2}$
$g'(t) = 4t^2 - 2 + \frac{t}{2}, t \in [0,1]$
Now, you have a quadratic on the right, whose positive root lies between 0 and 1, let this root be $t_{0}$. You can evaluate it but it involves $\sqrt(129)$
So, $g(t)$ is decreasing from 0 to $t_{0}$ and increasing from $t_{0}$ to 1.
So, $g(x)$ is increasing from $0$ to $cos^{-1}(t_{0})$ and decreasing from $cos^{-1}(t_{0})$ to $0$, so max value occurs at $x = cos^{-1}(t_{0})$.
Unfortunately, at this point, you do need to evaluate $t_{0}$ as $\frac{-1 + \sqrt(129)}{16}$, which can be approximated as 0.62, if you take $\sqrt(129)$ as 11. So, $cos(x) = 0.62$ and $sin(x) = 0.78$.
This makes the RHS of the AM-GM inequality as $\frac{(2*0.62*0.78 + 0.39 + 1)}{2}$, which becomes 1.1786. Square of which is less than 1.414.
The approximations are sufficiently small to not have an impact on the final result, I think since they are in the order of $10^{-2}$.
A: $$\sin x\cos x(2+\sin x)=\frac12\sin 2x \cdot (2+\sin x)=\sin 2x \cdot \left(1+\frac12\sin x\right)$$
$0\le\sin2x\le 1$, $\  \ 0\le\sin x\le 1$ for $x\in[0, \pi/2]$
$$0\le\sin (2x)\left(1+\frac12\sin x\right)<\sqrt2$$
$\therefore \sin x\cos x(2+\sin x)=\sqrt2$ has no solutions.
A: You can check that the only positive, hence feasible, root of the cubic is at $\sin x=t\approx0.748$ (I got this using Mathematica, but this is achievable rigorously using e.g. the cubic formula). At that value of $x$, we then have $\cos x\approx0.663$, so we bound
$$f(x)=\sin x\cdot \cos x\cdot(2+\sin x)<0.75\cdot0.665\cdot2.75<1.4<\sqrt{2}$$
and since this is the maximum the desired result is true.
