1. **State the problem:** Solve the equation $\sin x = \sin 282^\circ$ for $0^\circ \leq x \leq 270^\circ$. 2. **Recall the sine identity:** For angles $x$ and $a$, $\sin x = \sin a$ implies two possible solutions within one period: $$x = a + 360^\circ k \quad \text{or} \quad x = 180^\circ - a + 360^\circ k$$ where $k$ is any integer. 3. **Apply the identity with $a = 282^\circ$:** $$x = 282^\circ + 360^\circ k \quad \text{or} \quad x = 180^\circ - 282^\circ + 360^\circ k = -102^\circ + 360^\circ k$$ 4. **Find solutions within $0^\circ \leq x \leq 270^\circ$: - For $k=0$, $x = 282^\circ$ (not in the interval since $282^\circ > 270^\circ$) - For $k=0$, $x = -102^\circ$ (not in the interval since negative) - For $k=1$, $x = 282^\circ + 360^\circ = 642^\circ$ (too large) - For $k=1$, $x = -102^\circ + 360^\circ = 258^\circ$ (within the interval) 5. **Check the valid solution:** $x = 258^\circ$ satisfies $\sin x = \sin 282^\circ$ and lies in $0^\circ \leq x \leq 270^\circ$. **Final answer:** $$x = 258^\circ$$