The Position of the 2 Matters

Use $\sin^2 x=\frac{1-\cos 2x}2$ so that expression is:

$\frac{21}2+\color{#D61F06}{10\sin 2x-\frac{21\cos 2x}2 }$

$\frac{21}2+\color{#D61F06}{\frac{29}2\left[\sin \left(2x-\alpha\right) \right]}~~\small{(\alpha=\tan^{-1}(21/20)}$

Max value of red expression is $\frac{29}2$ when $2x-\alpha$ takes values $(4m+1)\frac{\pi}2$ for integral m. Hence max value of whole expression is $\frac{21+29}2=\boxed{25}$ .

$21\sin^2x+10\sin(2x)$

= $21\sin^2x+20\sin x\cos x$

= $21\sin^2x+4\sin^2x+20\sin x\cos x+4\cos^2x-4$

= $25\sin^2x+20\sin x\cos x+4\cos^2x-4$

= $(5\sin x+2\cos x)^2-4$

Now, we need to find the maximum possible value of $(5\sin x+2\cos x)^2$ .

By Cauchy-Schwarz Inequality,

$(25+4)(\sin^2x+\cos^2x)\geq(5\sin x+2\cos x)^2$

$(5\sin x+2\cos x)^2\leq 29$

Hence, the maximum possible value of $21\sin^2x+10\sin(2x)$ is $29-4=25$

Checking for the equality case, the equality holds when

$\frac{5}{\sin x}=\frac{2}{\cos x}$

This resolves to $\tan x = \frac{5}{2}$ , which can be achieved when $x$ is real.