How to solve for $\theta$ in $130\cos(\theta) - 122\sin(\theta)=10$?

Question

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?

Answer 1

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 $$ \theta=\arctan\frac{-14+\sqrt{271}}{30}+k\pi \qquad\text{or}\qquad \theta=\arctan\frac{-14-\sqrt{271}}{30}+k\pi $$

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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 $b+c-d=0$.

Answer 2

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 $$\sin(\phi+x)=\frac c{\sqrt {a^2+b^2}}$$

Answer 3

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 $\sin (A + B)$.