Task B-3.6. Solve the equation

$$
\left(\frac{\sin 1+\sin 3+\sin 5+\sin 7}{\cos 1+\cos 3+\cos 5+\cos 7}\right)^{\frac{1}{19 x}-107}=\frac{1}{\tan 4}
$$
## Solution.

By grouping conveniently and applying transformation formulas, we simplify the base on the left side of the equation.

$$
\begin{aligned}
& \frac{\sin 1 + \sin 3 + \sin 5 + \sin 7}{\cos 1 + \cos 3 + \cos 5 + \cos 7} = \\
= & \frac{(\sin 1 + \sin 7) + (\sin 3 + \sin 5)}{(\cos 1 + \cos 7) + (\cos 3 + \cos 5)} = \\
= & \frac{2 \sin \frac{1+7}{2} \cos \frac{1-7}{2} + 2 \sin \frac{3+5}{2} \cos \frac{3-5}{2}}{2 \cos \frac{1+7}{2} \cos \frac{1-7}{2} + 2 \cos \frac{3+5}{2} \cos \frac{3-5}{2}} = \\
= & \frac{2 \sin 4 \cos 3 + 2 \sin 4 \cos 1}{2 \cos 4 \cos 3 + 2 \cos 4 \cos 1} = \\
= & \frac{2 \sin 4 (\cos 3 + \cos 1)}{2 \cos 4 (\cos 3 + \cos 1)} = \tan 4.
\end{aligned}
$$

The given equation transforms into

$$
(\tan 4)^{\frac{1}{19 x} - 107} = \frac{1}{\tan 4}
$$

It follows that

$$
\begin{aligned}
& \frac{1}{19 x} - 107 = -1 \\
& \frac{1}{19 x} = 106 \\
& 2014 x = 1 \\
& x = \frac{1}{2014}
\end{aligned}
$$