Question

solve each equation on the interval ( 0 leq x leq 2pi ).

1. ( 3sin^2 x - cos^2 x = 0 )
2. ( sin 2x = 2sin x )

Explanation:

Step1: Use Pythagorean identity

$\sin^2 x = 1 - \cos^2 x$, substitute into equation 19:
$3(1 - \cos^2 x) - \cos^2 x = 0$

Step2: Simplify the equation

Expand and combine like terms:
$3 - 3\cos^2 x - \cos^2 x = 0$
$3 - 4\cos^2 x = 0$

Step3: Solve for $\cos x$

Rearrange and isolate $\cos^2 x$:
$4\cos^2 x = 3$
$\cos^2 x = \frac{3}{4}$
$\cos x = \pm\frac{\sqrt{3}}{2}$

Step4: Find $x$ for equation 19

For $\cos x = \frac{\sqrt{3}}{2}$ on $0 \leq x \leq 2\pi$:
$x = \frac{\pi}{6}, \frac{11\pi}{6}$
For $\cos x = -\frac{\sqrt{3}}{2}$ on $0 \leq x \leq 2\pi$:
$x = \frac{5\pi}{6}, \frac{7\pi}{6}$

Step5: Use double-angle identity for equation 20

$\sin 2x = 2\sin x \cos x$, substitute into equation 20:
$2\sin x \cos x = 2\sin x$

Step6: Rearrange and factor the equation

Subtract $2\sin x$ from both sides:
$2\sin x \cos x - 2\sin x = 0$
Factor out $2\sin x$:
$2\sin x(\cos x - 1) = 0$

Step7: Solve for $x$ for equation 20

Set each factor to 0:

1. $2\sin x = 0 \implies \sin x = 0$, so $x = 0, \pi, 2\pi$
2. $\cos x - 1 = 0 \implies \cos x = 1$, so $x = 0, 2\pi$

Combine unique solutions: $x = 0, \pi, 2\pi$

Answer:

1. For $3\sin^2 x - \cos^2 x = 0$: $x = \frac{\pi}{6}, \frac{5\pi}{6}, \frac{7\pi}{6}, \frac{11\pi}{6}$
2. For $\sin 2x = 2\sin x$: $x = 0, \pi, 2\pi$