Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
 A: $$3\sin x+4\cos x=A$$
$$3\cos x-4\sin x=2$$
$$9\sin^2x+24\sin x\cos x+16\cos^2x=A^2$$
$$9\cos^2x-24\sin x\cos x+16\sin^2x=4$$
add both equations
$$9+16=A^2+4$$
$$A^2=21$$
$$A=\pm \sqrt{21}$$
A: Think of it as a rotation equation. This is equivalent to the matrix formulation $$\begin{bmatrix} \cos x & -\sin x \\ \sin x & \cos x\end{bmatrix} \begin{bmatrix} 3 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ y \end{bmatrix}$$ where y is the unknown. Since the magnitude of a vector is preserved in rotation, $2^2 + y^2 = 3^2 + 4^2$ So, $y = \pm \sqrt {21}$.
A: $$a \cos(x) - b \sin(x) = \sqrt{a^2+b^2} \cos(x+y)$$ where $\cos(y) = a/\sqrt{a^2+b^2}$ and $\sin(y) = b/\sqrt{a^2 + b^2}$.  Then $$a \sin(x) + b \cos(x) = \sqrt{a^2 + b^2} \sin(x+y)$$
In your case, with $a=3$ and $b=4$, $\sqrt{a^2+b^2}=5$, $\cos(x+y) = 2/5$ so $\sin(x+y) = \pm \sqrt{1-(2/5)^2} = \pm  \sqrt{21}/5$, so the answer is $\pm \sqrt{21}$. 
A: This is really the same as aid78's method.
$$(3+4i)e^{ix}=(3\cos x-4\sin x)+i(3\sin x+4\cos x)
=2+iA$$
for $A=3\sin x+4\cos x$. As $|(3+4i)e^{ix}|^2=3^2+4^2=25$
then $25=4+A^2$, so $A=\pm\sqrt{21}$ (both are possible).
A: Hint:
Using Brahmagupta-Fibonacci Identity,
$$(a\cos x-b\sin x)^2+(a\sin x+b\cos x)^2=(a^2+b^2)(\cos^2x+\sin^2x)=?$$
