\( 8.5 \sin \left(\theta-20^{\circ}\right)=\cos 3 \theta \). Find the general solution.
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To solve the equation \( 8.5 \sin \left(\theta-20^{\circ}\right)=\cos 3 \theta \), we first express both sides in a suitable format. We know that \( \cos 3\theta = \sin \left(90^\circ - 3\theta\right) \), allowing us to write: \[ 8.5 \sin(\theta - 20^\circ) = \sin \left(90^\circ - 3\theta\right) \] Next, we can use the sine identity, which implies the equality: \[ \theta - 20^\circ = n\pi + (-1)^n \left(90^\circ - 3\theta\right) \] for \( n \in \mathbb{Z} \). This expands to two cases: 1. \( \theta - 20^\circ = n\pi + (90^\circ - 3\theta) \) 2. \( \theta - 20^\circ = n\pi - (90^\circ - 3\theta) \) ### Case 1: \[ \theta - 20^\circ = n\pi + 90^\circ - 3\theta \] Rearranging gives: \[ \theta + 3\theta = n\pi + 110^\circ \] \[ 4\theta = n\pi + 110^\circ \] \[ \theta = \frac{n\pi + 110^\circ}{4} \] ### Case 2: \[ \theta - 20^\circ = n\pi - (90^\circ - 3\theta) \] Rearranging gives: \[ \theta - 20^\circ = n\pi - 90^\circ + 3\theta \] \[ \theta - 3\theta = n\pi - 90^\circ + 20^\circ \] \[ -2\theta = n\pi - 70^\circ \] \[ \theta = -\frac{n\pi - 70^\circ}{2} \] or \[ \theta = \frac{70^\circ - n\pi}{2} \] Both expressions give us the general solutions. Therefore, the complete general solution is: \[ \theta = \frac{n\pi + 110^\circ}{4} \quad \text{and} \quad \theta = \frac{70^\circ - n\pi}{2}, \quad n \in \mathbb{Z} \]