Task B-4.4. Determine the value of the sine of the number, whose cosine is equal to its tangent.
## Solution.

If $\cos x = \tan x$, then

$$
\cos x = \frac{\sin x}{\cos x}, \quad x \neq \frac{\pi}{2} + k \pi, k \in \mathbb{Z}
$$

It follows that

$$
\begin{gathered}
\cos^2 x = \sin x \\
1 - \sin^2 x = \sin x \text{ or } \sin^2 x + \sin x - 1 = 0
\end{gathered}
$$

Then we have

$$
\sin x = t, \quad t^2 + t - 1 = 0
$$

The solutions to the quadratic equation are

$$
t_{1,2} = \frac{-1 \pm \sqrt{5}}{2}
$$

Since $\sin x \in [-1, 1]$, the only solution is

$$
\sin x = \frac{-1 + \sqrt{5}}{2}
$$