Anyone can help me with this system?
$$(I)\ 2x\sin(\theta)\cos(\theta) - 4x\sin(\theta) + 10\sin(\theta) = 0$$
$$(II)\ 10x\cos(\theta) - 2x^2\cos(\theta) + x^2(\cos^2(\theta) - \sin^2(\theta)) = 0$$
I tried to multiply (I) and (II) by several functions, but I did not come up with anything concrete.
There are two semesters I solved in a list of exercises, but I forgot how I did at the time.
The first equation is $2\sin\theta(x\cos\theta-2x+5)=0$
So, $\sin\theta=0$ or $x=\dfrac5{2-\cos\theta}$.
The second equation is $x[10\cos\theta-2x\cos\theta+x(\cos^2\theta-\sin^2\theta)]=0$
If $\sin\theta =0$, then $\cos\theta=\pm1$ and hence $x[\pm(10-2x)+x]=0$. This gives $x=10$ or $x=\dfrac{10}3$.
If $x=\dfrac{5}{2-\cos\theta}$, then $x\ne0$ and hence $10\cos\theta-2x\cos\theta+x(\cos^2\theta-\sin^2\theta)=0$.
\begin{align*} 10\cos\theta+\left(\dfrac{5}{2-\cos\theta}\right)(\cos^2\theta-\sin^2\theta-2\cos\theta)&=0\\ 10\cos\theta(2-\cos\theta)+5(\cos^2\theta-1+\cos^2\theta-2\cos\theta)&=0\\ \cos\theta&=\frac{1}{2}\\ x&=\frac{5}{2-\dfrac12}\\ x&=\frac{10}{3} \end{align*}