Problem source

Show that if  $\alpha$ is a solution of the equation
$$ 
5{\cos x} + 12{\sin x} = 7,
$$
then either
$$ 
{\cos  }{\alpha} = \frac{35 -12\sqrt{120}}{169}
$$
or $\cos \alpha$ has one other value which you should find.
 
Prove carefully that if 
$\frac{1}{2}\pi< \alpha < \pi$, then $\alpha < \frac{3}{4}\pi$.

Solution source

\begin{align*}
&& 5 \cos x + 12\sin x &= 7 \\
\Rightarrow && 5 \cos x - 7 &= -12 \sin x \\
\Rightarrow && 25 \cos^2 x - 70\cos x + 49 &= 144 \sin^2 x \\
\Rightarrow &&  25 \cos^2 x - 70\cos x + 49 &= 144 (1-\cos^2 x) \\
\Rightarrow && 169 \cos^2 x - 70 \cos x -95 &= 0 \\
\Rightarrow && \cos \alpha &= \frac{70 \pm \sqrt{70^2 - 4 \cdot 169 \cdot (-95)}}{2 \cdot 169} \\
&&&= \frac{35 \pm \sqrt{35^2 + 169 \cdot 95} }{169} \\
&&&= \frac{35 \pm 12\sqrt{120}}{169}
\end{align*}

If $\frac12 \pi < \alpha < \pi$ then $\cos \alpha$ is negative, in particular $\cos \alpha = \frac{35 -12\sqrt{120}}{169}$. Since $\cos$ is decreasing over this range, if $\cos \alpha > \cos \frac34 \pi = -\frac{\sqrt{2}}2$, then we will have shown $\alpha < \frac34 \pi$

\begin{align*}
&& \cos \alpha &= \frac{35 - 12 \sqrt{120}}{169} \\
&&&> \frac{35 - 12 \cdot \sqrt{121}}{169} \\
&&&= \frac{35 - 12 \cdot 11}{169} \\
&&&= \frac{35 - 132}{169} \\
&&&= -\frac{97}{169} \\
&&&> -\frac{8}{13}  
\end{align*}

but $\left ( \frac{8}{13} \right)^2 = \frac{64}{169} < \frac12$, so we are done.