Just Another Regular Trig Problem

$\begin{aligned} Q & = \frac {\sin x + \sin 2x}{\cos x + \cos 2x} & \small \color{#3D99F6} \text{Let }t = \tan \frac x2 \implies \sin x = \frac {2t}{1+t^2} \\ & = \frac {\frac {2t}{1+t^2}+2\cdot \frac {2t}{1+t^2}\cdot \frac {1-t^2}{1+t^2}}{\frac {1-t^2}{1+t^2}+1-2\left(\frac {2t}{1+t^2} \right)^2} & \small \color{#3D99F6} \implies \cos x = \frac {1-t^2}{1+t^2} \\ & = \frac {3t-t^3}{1-3t^2} \\ & = \frac {3\tan \frac x2 - \tan^3 \frac x2}{1-3 \tan^2 \frac x2} \\ & = \tan \frac 32 x \end{aligned}$

Therefore, $k = \dfrac 32 = \boxed {1.5}$ .

Use the sum to product formulas.

$\dfrac{\sin{x}+\sin{2x}}{\cos{x}+\cos{2x}}=\dfrac{2\sin{\frac{3x}{2}}\cos{\frac{-x}{2}}}{2\cos{\frac{3x}{2}}\cos{\frac{-x}{2}}}=\tan{\frac{3x}{2}}\implies k=1.5$