A calculus problem by A Former Brilliant Member

Multiply through by $\sin(x)$ to get the equation $\cos(x) + 1 = 5\sin(x)$ . Next, square both sides to find that

$\cos^{2}(x) + 2\cos(x) + 1 = 25\sin^{2}(x) = 25(1 - \cos^{2}(x))$

$\Longrightarrow 26\cos^{2}(x) + 2\cos(x) - 24 = 0 \Longrightarrow 13\cos^{2}(x) + \cos(x) - 12 = 0 \Longrightarrow (13\cos(x) - 12)(\cos(x) + 1) = 0$ .

So either $\cos(x) = \frac{12}{13}$ or $\cos(x) = -1$ . However, neither $\cot(x)$ nor $\csc(x)$ are defined when $\cos(x) = -1$ , so we are left with the unique solution $\cos(x) = \frac{12}{13} = \boxed{0.923}$ .

Check: With $x$ in the first quadrant, If $\cos(x) = \frac{12}{13}$ then $\csc(x) = \frac{13}{5}$ and $\cot(x) = \frac{12}{5}$ , giving us $\cot(x) + \csc(x) = 5$ as expected.