solve:- \( \left(0^{\circ} \leq \theta \leq 180^{\circ}\right) \) a) \( \sin 4 \theta+\sin 2 \theta=0 \)

To solve the equation \( \sin 4\theta + \sin 2\theta = 0 \), we can use the fact that \( \sin 4\theta = 2 \sin 2\theta \cos 2\theta \). This transforms our equation into: \[ 2\sin 2\theta \cos 2\theta + \sin 2\theta = 0 \] Factoring out \( \sin 2\theta \) gives: \[ \sin 2\theta (2\cos 2\theta + 1) = 0 \] Setting each factor to zero, we have: 1. \( \sin 2\theta = 0 \) 2. \( 2\cos 2\theta + 1 = 0 \) For \( \sin 2\theta = 0\): - This implies \( 2\theta = n\pi \), hence \( \theta = \frac{n\pi}{2} \). - Within the interval \( \left[ 0^{\circ}, 180^{\circ} \right] \), \( n = 0, 1, 2 \), giving \( \theta = 0^{\circ}, 90^{\circ}, 180^{\circ} \). For \( 2\cos 2\theta + 1 = 0 \): - Rearranging gives \( \cos 2\theta = -\frac{1}{2} \). - This occurs at \( 2\theta = 120^{\circ}, 240^{\circ} \) (plus \( 360^{\circ}k \) for integer \( k \)). - Dividing by 2, we find \( \theta = 60^{\circ}, 120^{\circ} \). Combining all solutions, the angles \( \theta \) that satisfy the original equation in the given range are: \( 0^{\circ}, 60^{\circ}, 90^{\circ}, 120^{\circ}, 180^{\circ} \).