How to solve for $\theta$ in $130\cos(\theta) - 122\sin(\theta)=10$? The original question was solve for $\theta$ in $65\cos(2\theta)-56\sin(2\theta)-55=0$. 
I reduced it to $\cos(\theta)((130\cos(\theta)-122\sin(\theta))-10=0$, therefore we have a $\theta=0$ but I don't know how to solve the above. 
Please could anyone help me solve it? 
 A: Hint: 
$$a \cos x+ b \sin x=c$$
$$\sqrt {a^2+b^2}(\frac a{\sqrt {a^2+b^2}} \cos x+ \frac b{\sqrt {a^2+b^2}} \sin x)=c$$
Let $\frac a{\sqrt {a^2+b^2}}=\sin \phi, \frac b{\sqrt {a^2+b^2}}=\cos \phi$. Then
$$\sin(\phi+x)=\frac c{\sqrt {a^2+b^2}}$$
A: Hint: You can write $A \cos 2x + B \sin 2x$ as $\sqrt{A^2+B^2} (\frac{A}{\sqrt{A^2+B^2}} \cos 2x + \frac{B}{\sqrt{A^2+B^2}} \sin 2x)$, then relate $\cos A \sin B + \sin A \cos B$ to $\sin (A + B)$.
A: Remember the formulas
$$
\cos2\theta=\frac{1-\tan^2\theta}{1+\tan^2\theta},
\qquad
\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}
$$
but first examine the cases $\theta=\pi/2$ and $\theta=-\pi/2$ that would invalidate the substitution.
We have
$$
65\cos\pi-56\sin\pi-55=-120\ne0\\
65\cos(-\pi)-56\sin(-\pi)-55=-120\ne0
$$
so the substitution is good and doesn't discard solutions.
Set $t=\tan\theta$ for simplicity, so you get
$$
65\frac{1-t^2}{1+t^2}-56\frac{2t}{1+t^2}-55=0
$$
that becomes
$$
60t^2+56t-5=0
$$
and the quadratic has roots
$$
\frac{-14+\sqrt{271}}{30}
\qquad
\frac{-14-\sqrt{271}}{30}
$$
so you get
$$
\theta=\arctan\frac{-14+\sqrt{271}}{30}+k\pi
\qquad\text{or}\qquad
\theta=\arctan\frac{-14-\sqrt{271}}{30}+k\pi
$$

Note also that your transformation to
$$
\cos\theta(130\cos\theta-122\sin\theta)-10=0
$$
does not reduce the equation to $130\cos\theta-122\sin\theta-10=0$. From
$$
a(b+c)-d=0
$$
you can't deduce $b+c-d=0$.
