<s>[INST] Show that $(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$ for some integers $A,B,C$ How do you solve this question?:$$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ is for all $x$. Here $A$, $B$ and $C$ is constants. I need to know $A$, $B$ and $C$ to pass this. They are  integers. 
I got this far:
(LS = left side)
$$LS =  49\cos^2(x)+\sin^2(x)+14\sin(x)\cos(x)$$
And then i follow some steps online and got right side to
$$RS = (A+C)\cos^2(x)+(C-A)\sin^2(x)+2B\sin(x)\cos(x)$$
Only problem is that i get C to 49/2.
 [/INST] $$(7 \cos{(x)}-\sin{(x)})^2=A  \cos{(2x)}+B \sin{(2x)}+C \\ \Rightarrow 49 \cos^2{(x)}-14 \cos{(x)} \sin{(x)}+\sin^2{(x)}=A \cos{(2x)}+B \sin{(2x)}+C \ \ \ (*)$$
Use the formulas:
$$\cos{(2x)}=\cos^2{(x)}-\sin^2{(x)} \text{ and } \sin{(2x)}=2 \sin{(x)} \cos{(x)}$$
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