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.
$$(7 \cos{(x)}-\sin{(x)})^2=A \cos{(2x)}+B \sin{(2x)}+C \\ \Rightarrow 49 \cos^2{(x)}-14 \cos{(x)} \sin{(x)}+\sin^2{(x)}=A \cos{(2x)}+B \sin{(2x)}+C \ \ \ (*)$$
Use the formulas:
$$\cos{(2x)}=\cos^2{(x)}-\sin^2{(x)} \text{ and } \sin{(2x)}=2 \sin{(x)} \cos{(x)}$$
apply the following identities to your LHS:
And don't touch the original RHS!
Second method: you can give three different values to $x$ and solve a linear system: for example set $x=0$ ----> $A+C=49$
$x=\pi/4$ ----->?
Anything forall x implies an identity so you can put any * values of x to get C. for e.g.: $$(7\cos(x)-\sin(x))^2=A\cos(2x)+B\sin(2x)+C$$ $$ \left\{ \begin{array}{c} 49=A+C\quad x=0 \\ 1=-A+C\quad x=\pi/2\\ 18=B+C\\ \end{array} \right. $$ Solving these we get: $$ \left\{ \begin{array}{c} A=24 \\ C=25\\ B=-7\\ \end{array} \right. $$
Alternatively: $$(7\cos(x)-\sin(x))^2=49\cos^2x+\sin^2x-14\cos x\sin x \\=49\left(\frac{1+\cos{2x}}2\right)+\left(\frac{1-\cos{2x}}2\right)-7\sin 2x \\=24\cos 2x-7\sin 2x+25$$ So $A=24,B=-7,C=25$
Use: $$\cos 2x=2\cos^2x-1=1-2\sin^2x,\sin 2x=2\sin x\cos x$$
To match coefficients of $\cos^2 x$ and of $\sin^2 x$ you were solving $$ \left\{ \begin{array}{c} C+A = 49 \\ C-A = 1 \end{array} \right. $$ This gives $C = 25$ and $A = 24$.
You probably did not notice that $1 \cdot \sin^2 x$ was needed so you used a zero in the second equation, getting $C = A = 49/2$.
