Solve for $x$ and $y$, The equations are $x \cos^{3} y+3x \cos y \sin^{2} y =14 $ and $ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $ Consider the system of equations 
$$x \cos^{3} y+3x \cos y \sin^{2} y =14 $$
$$ x \sin^{3} y+3x \cos^{2} y \sin y = 13 $$
$1)$ the values of $x$ is /are..
Answer is $ \pm\sqrt 5 $
The number of values of $y$ in $(0,6 \pi)$is 
Answer is $6$
$Sin^2 y + 2cos^2 y $ is  
Answer is $ \frac95 $
I added both the equation and make whole cube 
I subtracted both the equation and make whole cube then I divided these equations to  find the value of $ \tan y $ = $ \dfrac{1}{2} $
I divided the first equation with $ cos^2 y $ and plugged the value of $\tan y$ which gives $x= +5\sqrt 5 $. But I also need -$ 5\sqrt 5 $ I have no idea where I go wrong.
 A: I think your approach is quite good.
\begin{align*}
\frac{x(\sin^3y+3\cos^2y\sin y)}{x(\cos^3y+3\cos y\sin^2y)}&=\frac{13}{14}\\
\frac{\tan^3y+3\tan y}{1+3\tan^2y}&=\frac{13}{14}\\
14\tan^3y-39\tan^2y+42\tan y-13&=0\\
(2\tan y-1)(7\tan^2y-16\tan y+13)&=0
\end{align*}
$\tan y=\frac12$.
$y$ has $6$ values in $(0,6\pi)$.
$\sec^2y=1+(\frac12)^2=\frac{5}{4}$.
$\cos y=\pm\sqrt{\frac45}=\pm\frac{2}{\sqrt5}$.
$\sin y=\tan y\cos y=\pm\frac{1}{\sqrt5}$.
So, we have
\begin{align*}
\pm x\left[\left(\frac{1}{\sqrt5}\right)^3+3\left(\frac{2}{\sqrt5}\right)^2\left(\frac{1}{\sqrt5}\right)\right]&=13\\
x&=\pm5\sqrt5
\end{align*}
$\sin^2y+2\cos^2y=(\frac{1}{\sqrt5})^2+2(\frac{2}{\sqrt5})^2=\frac95$.
A: When you render $\tan y =1/2$, it follows that
$1+\tan^2 y = \sec^2 y = 1/\cos^2 y = 5/4$
But then you have to allow both the positive and negative square roots for the cosine.  One period of the cosine is $2\pi$ which is two periods of the tangent function, so both the positive and negative cosine values occur in different periods of the tangent function.  When you allow the negative value, $\cos y = -2/\sqrt{5}$, you pick up the negative value for $x$.
