16. Algebraically solve the following equations for the given domain.
a) $\cos 2 x=1-2 \sin x, 0 \leq x \leq 2 \pi$
b) $\sin 2 x+\cos x=0,-180^{\circ} \leq x \leq 180^{\circ}$
Answer
Final Answer: The solutions to the equation \(\cos 2 x=1-2 \sin x, 0 \leq x \leq 2 \pi\) within the given domain are \(x = 0, \frac{\pi}{2}, \pi\). So, the final answer is \(\boxed{0, \frac{\pi}{2}, \pi}\).
Step 1 ：We are given the equation \(\cos 2x = 1 - 2\sin x\) with the domain \(0 \leq x \leq 2 \pi\).
Step 2 ：First, we use the double angle identity for cosine, which is \(\cos 2x = 1 - 2\sin^2x\).
Step 3 ：We then set this equal to the right side of the equation and solve for \(\sin x\).
Step 4 ：The solutions to the equation are \(x = 0, \frac{\pi}{2}, \pi\).
Step 5 ：We need to check if these solutions are within the given domain \(0 \leq x \leq 2 \pi\).
Step 6 ：The solutions \(x = 0, \frac{\pi}{2}, \pi\) are indeed within the given domain.
Step 7 ：Final Answer: The solutions to the equation \(\cos 2 x=1-2 \sin x, 0 \leq x \leq 2 \pi\) within the given domain are \(x = 0, \frac{\pi}{2}, \pi\). So, the final answer is \(\boxed{0, \frac{\pi}{2}, \pi}\).
