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

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.

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 $$\sin(2x)+\cos(x)-\cos^3(x)\leq\sqrt2.$$

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.