1. **State the problem:** We need to sketch the graphs of $y = \sin 2x$ and $y = 1 + \cos 2x$ for $0^\circ \leq x \leq 360^\circ$ and find the number of solutions to the equation $\sin 2x = 1 + \cos 2x$ in this interval. 2. **Recall the functions and their properties:** - $y = \sin 2x$ is a sine wave with frequency doubled, so it completes two full cycles from $0^\circ$ to $360^\circ$. - $y = 1 + \cos 2x$ is a cosine wave with frequency doubled and shifted up by 1 unit. 3. **Sketching the graphs:** - $\sin 2x$ oscillates between $-1$ and $1$ with zeros at $x=0^\circ, 90^\circ, 180^\circ, 270^\circ, 360^\circ$. - $1 + \cos 2x$ oscillates between $0$ and $2$ with maxima at $x=0^\circ, 180^\circ, 360^\circ$ and minima at $x=90^\circ, 270^\circ$. 4. **Find the number of solutions to $\sin 2x = 1 + \cos 2x$:** Rewrite the equation: $$\sin 2x = 1 + \cos 2x$$ Use the identity $\sin 2x - \cos 2x = 1$. 5. **Express $\sin 2x - \cos 2x$ as a single sine function:** Recall that: $$\sin A - \cos A = \sqrt{2} \sin \left(A - 45^\circ\right)$$ So, $$\sin 2x - \cos 2x = \sqrt{2} \sin \left(2x - 45^\circ\right)$$ 6. **Set the equation:** $$\sqrt{2} \sin \left(2x - 45^\circ\right) = 1$$ Divide both sides by $\sqrt{2}$: $$\sin \left(2x - 45^\circ\right) = \frac{1}{\sqrt{2}} = \sin 45^\circ$$ 7. **Solve for $2x - 45^\circ$:** The general solutions for $\sin \theta = \sin \alpha$ are: $$\theta = \alpha + 360^\circ k \quad \text{or} \quad \theta = 180^\circ - \alpha + 360^\circ k$$ where $k$ is any integer. So, $$2x - 45^\circ = 45^\circ + 360^\circ k \quad \text{or} \quad 2x - 45^\circ = 135^\circ + 360^\circ k$$ 8. **Solve for $x$:** For the first case: $$2x = 90^\circ + 360^\circ k \Rightarrow x = 45^\circ + 180^\circ k$$ For the second case: $$2x = 180^\circ + 360^\circ k \Rightarrow x = 90^\circ + 180^\circ k$$ 9. **Find all solutions in $0^\circ \leq x \leq 360^\circ$:** For $x = 45^\circ + 180^\circ k$: - $k=0 \Rightarrow x=45^\circ$ - $k=1 \Rightarrow x=225^\circ$ - $k=2 \Rightarrow x=405^\circ$ (outside range) For $x = 90^\circ + 180^\circ k$: - $k=0 \Rightarrow x=90^\circ$ - $k=1 \Rightarrow x=270^\circ$ - $k=2 \Rightarrow x=450^\circ$ (outside range) 10. **Count the solutions:** The solutions within the interval are: $$45^\circ, 90^\circ, 225^\circ, 270^\circ$$ There are 4 solutions. **Final answer:** - The graphs of $y=\sin 2x$ and $y=1+\cos 2x$ intersect 4 times in $0^\circ \leq x \leq 360^\circ$.