Question

25-26 precalculus ic sem b
trigonometric double angle identities
solving a trigonometric equation
over the interval \\((0, 2\pi)\\), what are the solutions to \\(\sin(2x) = 2\cos(x)\\)? choose two correct answers.
options: \\(\pi\\), \\(\frac{\pi}{6}\\), \\(\frac{5\pi}{6}\\), \\(\frac{\pi}{2}\\), \\(\frac{11\pi}{6}\\), \\(\frac{3\pi}{2}\\) (note: the original image shows some of these options in boxes, but the text here captures the question and options)

Explanation:

Step1: Use double - angle identity

We know that the double - angle identity for sine is $\sin(2x)=2\sin x\cos x$. The given equation is $\sin(2x) = 2\cos(x)$. Substitute $\sin(2x)$ with $2\sin x\cos x$:
$$2\sin x\cos x=2\cos x$$

Step2: Rearrange the equation

Subtract $2\cos x$ from both sides of the equation:
$$2\sin x\cos x - 2\cos x=0$$
Factor out $2\cos x$ from the left - hand side:
$$2\cos x(\sin x - 1)=0$$

Step3: Solve for $\cos x$ and $\sin x$

Set each factor equal to zero:
- Case 1: $2\cos x = 0\Rightarrow\cos x=0$
- Case 2: $\sin x - 1=0\Rightarrow\sin x = 1$

Step4: Find solutions in the interval $(0,2\pi)$

- For $\cos x = 0$, in the interval $(0,2\pi)$, $x=\frac{\pi}{2}$ or $x = \frac{3\pi}{2}$. But we will check later if these are valid (we made a substitution, but we need to verify in the original equation).
- For $\sin x=1$, in the interval $(0,2\pi)$, $x = \frac{\pi}{2}$.

Now, let's check the original equation $\sin(2x)=2\cos(x)$ for the values:
- When $x = \frac{\pi}{6}$:
Left - hand side: $\sin(2\times\frac{\pi}{6})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$
Right - hand side: $2\cos(\frac{\pi}{6})=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$, $\frac{\sqrt{3}}{2}
eq\sqrt{3}$, so $x = \frac{\pi}{6}$ is not a solution.
- When $x=\pi$:
Left - hand side: $\sin(2\pi)=0$
Right - hand side: $2\cos(\pi)=- 2$, $0
eq - 2$, so $x = \pi$ is not a solution.
- When $x=\frac{5\pi}{6}$:
Left - hand side: $\sin(2\times\frac{5\pi}{6})=\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}$
Right - hand side: $2\cos(\frac{5\pi}{6})=2\times(-\frac{\sqrt{3}}{2})=-\sqrt{3}$, $-\frac{\sqrt{3}}{2}
eq-\sqrt{3}$, so $x=\frac{5\pi}{6}$ is not a solution.
- When $x = \frac{\pi}{2}$:
Left - hand side: $\sin(2\times\frac{\pi}{2})=\sin(\pi)=0$
Right - hand side: $2\cos(\frac{\pi}{2})=0$, so $x=\frac{\pi}{2}$ is a solution.
- When $x=\frac{11\pi}{6}$:
Left - hand side: $\sin(2\times\frac{11\pi}{6})=\sin(\frac{11\pi}{3})=\sin(\frac{11\pi}{3}-4\pi)=\sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$
Right - hand side: $2\cos(\frac{11\pi}{6})=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$, $-\frac{\sqrt{3}}{2}
eq\sqrt{3}$, so $x=\frac{11\pi}{6}$ is not a solution.
- When $x=\frac{3\pi}{2}$:
Left - hand side: $\sin(2\times\frac{3\pi}{2})=\sin(3\pi)=0$
Right - hand side: $2\cos(\frac{3\pi}{2})=0$, so $x = \frac{3\pi}{2}$ is a solution. Wait, but the options given are $\pi,\frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2},\frac{11\pi}{6},\frac{3\pi}{2}$. Maybe we made a mistake in the initial factoring. Let's go back.

Starting from $2\sin x\cos x-2\cos x = 0$, we can also divide both sides by $2\cos x$ (assuming $\cos x
eq0$):
If $\cos x
eq0$, then $2\sin x\cos x-2\cos x=0\Rightarrow2\cos x(\sin x - 1)=0\Rightarrow\sin x=1$ (when $\cos x
eq0$ is not valid here). Wait, another approach:

The original equation is $\sin(2x)=2\cos x$. Let's write all solutions:

From $\sin(2x)-2\cos x = 0\Rightarrow2\sin x\cos x-2\cos x=0\Rightarrow2\cos x(\sin x - 1)=0$

So either $\cos x = 0$ or $\sin x=1$.

For $\cos x = 0$, $x=\frac{\pi}{2},\frac{3\pi}{2}$

For $\sin x = 1$, $x=\frac{\pi}{2}$

Now, let's check the options:

- Option $\frac{\pi}{2}$: As we saw, $\sin(2\times\frac{\pi}{2})=\sin(\pi)=0$, $2\cos(\frac{\pi}{2}) = 0$, so it is a solution.
- Option $\frac{3\pi}{2}$: $\sin(2\times\frac{3\pi}{2})=\sin(3\pi)=0$, $2\cos(\frac{3\pi}{2})=0$, so it is a solution. But maybe the question has a different set of options. Wait, maybe we misapplied the identity. Wait, the original equation is $\sin(2x)=2\cos x$. Let's try $x = \frac{\pi}{6}$:

$\sin(2\times\frac{\pi}{6})=\sin(\frac{\pi}{3})=\frac{\sqrt{3}}{2}$, $2\cos(\frac{\pi}{6})=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$, not equal.

$x=\frac{5\pi}{6}$: $\sin(2\times\frac{5\pi}{6})=\sin(\frac{5\pi}{3})=-\frac{\sqrt{3}}{2}$, $2\cos(\frac{5\pi}{6})=2\times(-\frac{\sqrt{3}}{2})=-\sqrt{3}$, not equal.

$x=\pi$: $\sin(2\pi)=0$, $2\cos(\pi)=-2$, not equal.

$x=\frac{11\pi}{6}$: $\sin(2\times\frac{11\pi}{6})=\sin(\frac{11\pi}{3})=\sin(\frac{11\pi}{3}-4\pi)=\sin(-\frac{\pi}{3})=-\frac{\sqrt{3}}{2}$, $2\cos(\frac{11\pi}{6})=2\times\frac{\sqrt{3}}{2}=\sqrt{3}$, not equal.

Wait, maybe the question has a typo or we misread. Wait, let's re - solve the equation:

$\sin(2x)=2\cos x$

$2\sin x\cos x-2\cos x = 0$

$2\cos x(\sin x - 1)=0$

So $\cos x = 0$ or $\sin x=1$

$\cos x = 0\Rightarrow x=\frac{\pi}{2},\frac{3\pi}{2}$

$\sin x = 1\Rightarrow x=\frac{\pi}{2}$

So the solutions in the interval $(0,2\pi)$ are $x=\frac{\pi}{2}$ and $x=\frac{3\pi}{2}$ among the given options.

Answer:

$\frac{\pi}{2}$, $\frac{3\pi}{2}$