Find all the values of $\theta$, where $0^\circ \leq \theta \leq 360^\circ$, such that $$ \sin(\theta - 60^\circ) = \frac{\sqrt{3}}{2} - HSC - SSCE Mathematics Advanced - Question 20 - 2023 - Paper 1

By setting the equation as: $\theta - 60^\circ = 60^\circ$ we can solve for $\theta$: $\theta = 60^\circ + 60^\circ = 120^\circ.$

Additionally, since the sine function is positive in both the first and second quadrants, we can also set: $\theta - 60^\circ = 120^\circ$ This gives: $\theta = 120^\circ + 60^\circ = 180^\circ.$

Further, because sine is also positive for the angle $240^\circ$ in the third quadrant, we can also set: $\theta - 60^\circ = 240^\circ$ Solving gives: $\theta = 240^\circ + 60^\circ = 300^\circ.$

Thus, the final values of $\theta$ that satisfy the original equation are: $\theta = 120^\circ, 300^\circ, 360^\circ.$