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What are all the solutions to the equation [tex]\sin 2x = 2\cos x[/tex] in the interval [tex][0, 2\pi)[/tex]?

A. [tex]\{0, \pi\}[/tex]
B. [tex]\left\{\frac{\pi}{2}\right\}[/tex]
C. [tex]\left\{\frac{\pi}{2}, \frac{3\pi}{2}\right\}[/tex]


Sagot :

Sure, let's solve the equation [tex]\(\sin(2x) = 2\cos(x)\)[/tex] for [tex]\(x\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex].

### Step-by-Step Solution:

1. Rewrite the Equation:
The equation we are dealing with is:
[tex]\[ \sin(2x) = 2\cos(x) \][/tex]

Using the double-angle identity for sine, we can rewrite [tex]\(\sin(2x)\)[/tex] as [tex]\(2\sin(x)\cos(x)\)[/tex]:
[tex]\[ 2\sin(x)\cos(x) = 2\cos(x) \][/tex]

2. Isolate Terms:
Divide both sides of the equation by 2:
[tex]\[ \sin(x)\cos(x) = \cos(x) \][/tex]

3. Solve for [tex]\(\cos(x)\)[/tex]:
We can rearrange this equation to isolate [tex]\(\cos(x)\)[/tex]:
[tex]\[ \sin(x)\cos(x) - \cos(x) = 0 \][/tex]

Factor out [tex]\(\cos(x)\)[/tex]:
[tex]\[ \cos(x)(\sin(x) - 1) = 0 \][/tex]

4. Find the Zeros:
Now set each factor to zero and solve for [tex]\(x\)[/tex]:
[tex]\[ \cos(x) = 0 \quad \text{or} \quad \sin(x) - 1 = 0 \][/tex]

- For [tex]\(\cos(x) = 0\)[/tex]:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \quad \text{within the interval} \; [0, 2\pi) \][/tex]

- For [tex]\(\sin(x) = 1\)[/tex]:
[tex]\[ x = \frac{\pi}{2} \quad \text{within the interval} \; [0, 2\pi) \][/tex]

5. Combine Solutions:
From these calculations we find:
[tex]\[ x = \frac{\pi}{2}, \frac{3\pi}{2} \][/tex]

However, we need to verify that both of these values satisfy the original equation [tex]\(\sin(2x) = 2\cos(x)\)[/tex]. For [tex]\(x = \frac{\pi}{2}\)[/tex]:

[tex]\(\sin\left(2 \cdot \frac{\pi}{2}\right) = \sin(\pi) = 0\)[/tex]

[tex]\(2\cos\left(\frac{\pi}{2}\right) = 2 \cdot 0 = 0\)[/tex]

Therefore, [tex]\(x = \frac{\pi}{2}\)[/tex] is a solution.

For [tex]\(x = \frac{3\pi}{2}\)[/tex]:

[tex]\(\sin\left(2 \cdot \frac{3\pi}{2}\right) = \sin(3\pi) = 0\)[/tex]

[tex]\(2\cos\left(\frac{3\pi}{2}\right) = 2 \cdot 0 = 0\)[/tex]

Therefore, [tex]\(x = \frac{3\pi}{2}\)[/tex] is also a solution.

### Conclusion
The solutions to the equation [tex]\(\sin 2x = 2\cos x\)[/tex] in the interval [tex]\([0, 2\pi)\)[/tex] are:
[tex]\[ \left\{ \frac{\pi}{2}, \frac{3\pi}{2} \right\} \][/tex]
So, the correct answer is:
[tex]\[ \boxed{\left\{ \frac{\pi}{2}, \frac{3\pi}{2} \right\}} \][/tex]