1. **State the problem:** Find all values of $x$ such that $$\cos(2x) = -\frac{\sqrt{3}}{2}$$ for $$0^\circ \leq x \leq 360^\circ$$. 2. **Recall the cosine values:** The cosine function equals $$-\frac{\sqrt{3}}{2}$$ at angles $$120^\circ$$ and $$240^\circ$$ within one full rotation $$0^\circ \leq \theta < 360^\circ$$. 3. **Set the argument equal to these angles:** Since $$\cos(2x) = -\frac{\sqrt{3}}{2}$$, we have $$2x = 120^\circ + 360^\circ k \quad \text{or} \quad 2x = 240^\circ + 360^\circ k$$ where $$k$$ is any integer. 4. **Solve for $$x$$:** $$x = \frac{120^\circ + 360^\circ k}{2} = 60^\circ + 180^\circ k$$ $$x = \frac{240^\circ + 360^\circ k}{2} = 120^\circ + 180^\circ k$$ 5. **Find all $$x$$ in the interval $$0^\circ \leq x \leq 360^\circ$$:** - For $$x = 60^\circ + 180^\circ k$$: - $$k=0 \Rightarrow x=60^\circ$$ - $$k=1 \Rightarrow x=240^\circ$$ - $$k=2 \Rightarrow x=420^\circ$$ (outside interval) - For $$x = 120^\circ + 180^\circ k$$: - $$k=0 \Rightarrow x=120^\circ$$ - $$k=1 \Rightarrow x=300^\circ$$ - $$k=2 \Rightarrow x=480^\circ$$ (outside interval) 6. **Final solutions:** $$x = 60^\circ, 120^\circ, 240^\circ, 300^\circ$$. These are all the values of $$x$$ in the given interval satisfying the equation.