Questions: Find all angles, (0^circ leq theta<360^circ), that satisfy the equation below, to the nearest tenth of a degree (if necessary). [6 sin theta-6=sin theta-8]

Solution Steps

Start by moving all terms involving \(\sin \theta\) to one side and constants to the other side: \[ 6 \sin \theta - \sin \theta = -8 + 6 \] Simplify: \[ 5 \sin \theta = -2 \]

Divide both sides by 5 to isolate \(\sin \theta\): \[ \sin \theta = \frac{-2}{5} \] \[ \sin \theta = -0.4 \]

Calculate the reference angle \(\alpha\) using the inverse sine function: \[ \alpha = \sin^{-1}(0.4) \approx 23.6^{\circ} \]

Since \(\sin \theta\) is negative, the angles lie in the third and fourth quadrants: \[ \theta_1 = 180^{\circ} + 23.6^{\circ} = 203.6^{\circ} \] \[ \theta_2 = 360^{\circ} - 23.6^{\circ} = 336.4^{\circ} \]

Final Answer