2) \( \operatorname{tg} 20^{\circ}-4 \sin 20^{\circ} \cdot \sin 50^{\circ}=-2 \sin 20^{\circ} \)

To solve the equation \( \operatorname{tg} 20^{\circ}-4 \sin 20^{\circ} \cdot \sin 50^{\circ}=-2 \sin 20^{\circ} \), we can first rewrite the tangent function in terms of sine and cosine: \[ \operatorname{tg} 20^{\circ} = \frac{\sin 20^{\circ}}{\cos 20^{\circ}}. \] Substituting this into the equation gives: \[ \frac{\sin 20^{\circ}}{\cos 20^{\circ}} - 4 \sin 20^{\circ} \sin 50^{\circ} = -2 \sin 20^{\circ}. \] Now, let's factor out \( \sin 20^{\circ} \): \[ \sin 20^{\circ} \left(\frac{1}{\cos 20^{\circ}} - 4 \sin 50^{\circ} + 2\right) = 0. \] This leads to two cases: 1. \( \sin 20^{\circ} = 0 \) (not applicable, as \( 20^{\circ} \) is not an angle where sine is zero) 2. \( \frac{1}{\cos 20^{\circ}} - 4 \sin 50^{\circ} + 2 = 0.\) To simplify this, we find \( \sin 50^{\circ} \). With the identity \( \sin 50^{\circ} = \cos 40^{\circ} \): \[ \frac{1}{\cos 20^{\circ}} - 4 \cos 40^{\circ} + 2 = 0. \] From here, multiply through by \( \cos 20^{\circ} \): \[ 1 - 4 \cos 40^{\circ} \cos 20^{\circ} + 2 \cos 20^{\circ} = 0. \] With further manipulation, you can isolate \( \cos 20^{\circ} \) leading to a solvable quadratic in \( \cos 20^{\circ} \). Overall, simplifying and solving should yield the angle relationship you're looking for. Once you find \( \cos 20^{\circ} \) or \( \sin 20^{\circ} \), you can calculate the necessary values accordingly.