a) \( \frac{\cos ^{2} 60^{\circ}-\sin 30^{\circ}+\cos ^{2} 30^{\circ}}{\sin 60^{\circ} \sin x \operatorname{tg} 30^{\circ}+\sin x \sin 30^{\circ}}=1 \)

To solve the equation \[ \frac{\cos ^{2} 60^{\circ}-\sin 30^{\circ}+\cos ^{2} 30^{\circ}}{\sin 60^{\circ} \sin x \tan 30^{\circ}+\sin x \sin 30^{\circ}}=1, \] we first need to evaluate the trigonometric functions involved. 1. **Calculate the trigonometric values:** - \( \cos 60^{\circ} = \frac{1}{2} \) - \( \sin 30^{\circ} = \frac{1}{2} \) - \( \cos 30^{\circ} = \frac{\sqrt{3}}{2} \) - \( \sin 60^{\circ} = \frac{\sqrt{3}}{2} \) - \( \tan 30^{\circ} = \frac{1}{\sqrt{3}} \) 2. **Substituting these values into the equation:** - Calculate \( \cos^2 60^{\circ} \): \[ \cos^2 60^{\circ} = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \] - Calculate \( \cos^2 30^{\circ} \): \[ \cos^2 30^{\circ} = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \] - Substitute these into the numerator: \[ \cos^2 60^{\circ} - \sin 30^{\circ} + \cos^2 30^{\circ} = \frac{1}{4} - \frac{1}{2} + \frac{3}{4} \] Simplifying this: \[ = \frac{1}{4} - \frac{2}{4} + \frac{3}{4} = \frac{1}{4} + \frac{3}{4} - \frac{2}{4} = \frac{2}{4} = \frac{1}{2} \] 3. **Now calculate the denominator:** - Substitute the values into the denominator: \[ \sin 60^{\circ} \sin x \tan 30^{\circ} + \sin x \sin 30^{\circ} = \frac{\sqrt{3}}{2} \sin x \cdot \frac{1}{\sqrt{3}} + \sin x \cdot \frac{1}{2} \] Simplifying this: \[ = \frac{\sin x}{2} + \frac{\sin x}{2} = \sin x \] 4. **Now substitute back into the equation:** \[ \frac{\frac{1}{2}}{\sin x} = 1 \] 5. **Cross-multiply to solve for \(\sin x\):** \[ \frac{1}{2} = \sin x \] 6. **Find \(x\):** \[ \sin x = \frac{1}{2} \] The solutions for this equation are: \[ x = 30^{\circ} + 360^{\circ}k \quad \text{or} \quad x = 150^{\circ} + 360^{\circ}k \quad (k \in \mathbb{Z}) \] Thus, the final solutions for \(x\) are: \[ x = 30^{\circ} + 360^{\circ}k \quad \text{or} \quad x = 150^{\circ} + 360^{\circ}k \quad (k \in \mathbb{Z}). \]