Question

question 13
solve $2\sin(2\phi) - 2\cos(\phi) = 0$ for all solutions $0 \leq \phi < 2\pi$
$\phi = $
give your answers accurate to at least 2 decimal places, as a list separated by commas
add work
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Explanation:

Step1: Use Double-Angle Identity

Recall the double-angle identity for sine: $\sin(2\phi) = 2\sin(\phi)\cos(\phi)$. Substitute this into the equation:
$2(2\sin(\phi)\cos(\phi)) - 2\cos(\phi) = 0$
Simplify: $4\sin(\phi)\cos(\phi) - 2\cos(\phi) = 0$

Step2: Factor Out Common Term

Factor out $2\cos(\phi)$ from the left side:
$2\cos(\phi)(2\sin(\phi) - 1) = 0$

Step3: Solve Each Factor

Case 1: $\cos(\phi) = 0$

For $0 \leq \phi < 2\pi$, $\cos(\phi) = 0$ when $\phi = \frac{\pi}{2} \approx 1.57$ or $\phi = \frac{3\pi}{2} \approx 4.71$.

Case 2: $2\sin(\phi) - 1 = 0$

Solve for $\sin(\phi)$: $\sin(\phi) = \frac{1}{2}$
For $0 \leq \phi < 2\pi$, $\sin(\phi) = \frac{1}{2}$ when $\phi = \frac{\pi}{6} \approx 0.52$ or $\phi = \frac{5\pi}{6} \approx 2.62$.

Answer:

0.52, 1.57, 2.62, 4.71