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Math  /  Trigonometry

Question15. Solve 3sin⁡2x−1=03 \sin 2 x-1=0 on the interval x∈[0,2π]x \in[0,2 \pi].

Studdy Solution

STEP 1

1. The equation 3sin⁡2x−1=0 3 \sin 2x - 1 = 0 is a trigonometric equation.
2. We are solving for x x within the interval [0,2π] [0, 2\pi] .
3. The function sin⁡2x\sin 2x has a period of π\pi.

STEP 2

1. Isolate the trigonometric function sin⁡2x\sin 2x.
2. Solve the equation sin⁡2x=13\sin 2x = \frac{1}{3}.
3. Find all solutions for 2x2x within the interval [0,4π][0, 4\pi].
4. Solve for xx by dividing the solutions for 2x2x by 2.
5. Ensure all solutions for xx fall within the interval [0,2π][0, 2\pi].

STEP 3

First, isolate sin⁡2x\sin 2x by adding 1 to both sides and then dividing by 3:
3sin⁡2x−1=0 3 \sin 2x - 1 = 0 3sin⁡2x=1 3 \sin 2x = 1 sin⁡2x=13 \sin 2x = \frac{1}{3}

STEP 4

Solve the equation sin⁡2x=13\sin 2x = \frac{1}{3}. We need to find the general solutions for 2x2x:
The general solution for sin⁡θ=a\sin \theta = a is given by: θ=arcsin⁡(a)+2kπorθ=π−arcsin⁡(a)+2kπ \theta = \arcsin(a) + 2k\pi \quad \text{or} \quad \theta = \pi - \arcsin(a) + 2k\pi
For sin⁡2x=13\sin 2x = \frac{1}{3}, we have: 2x=arcsin⁡(13)+2kπ 2x = \arcsin\left(\frac{1}{3}\right) + 2k\pi 2x=π−arcsin⁡(13)+2kπ 2x = \pi - \arcsin\left(\frac{1}{3}\right) + 2k\pi

STEP 5

Find all solutions for 2x2x within the interval [0,4π][0, 4\pi]:
1. Calculate arcsin⁡(13)\arcsin\left(\frac{1}{3}\right). Let's denote it as α\alpha.
2. The solutions for 2x2x are: - 2x=α+2kπ2x = \alpha + 2k\pi - 2x=π−α+2kπ2x = \pi - \alpha + 2k\pi
For k=0k = 0: - 2x=α2x = \alpha - 2x=π−α2x = \pi - \alpha
For k=1k = 1: - 2x=α+2π2x = \alpha + 2\pi - 2x=π−α+2π2x = \pi - \alpha + 2\pi
Check which solutions fall within [0,4π][0, 4\pi].

STEP 6

Calculate specific values for 2x2x and check the interval:
1. 2x=α2x = \alpha
2. 2x=π−α2x = \pi - \alpha
3. 2x=α+2π2x = \alpha + 2\pi
4. 2x=π−α+2π2x = \pi - \alpha + 2\pi
Assuming α=arcsin⁡(13)\alpha = \arcsin\left(\frac{1}{3}\right), compute the approximate values and ensure they fall within [0,4π][0, 4\pi].

STEP 7

Solve for xx by dividing each valid 2x2x solution by 2:
1. x=α2x = \frac{\alpha}{2}
2. x=π−α2x = \frac{\pi - \alpha}{2}
3. x=α+2π2x = \frac{\alpha + 2\pi}{2}
4. x=π−α+2π2x = \frac{\pi - \alpha + 2\pi}{2}

STEP 8

Ensure all solutions for xx fall within the interval [0,2π][0, 2\pi]:
Verify each calculated xx value is within the interval [0,2π][0, 2\pi].
The solutions for xx are approximately:
x=arcsin⁡(13)2,π−arcsin⁡(13)2,arcsin⁡(13)+2π2,π−arcsin⁡(13)+2π2 x = \frac{\arcsin\left(\frac{1}{3}\right)}{2}, \frac{\pi - \arcsin\left(\frac{1}{3}\right)}{2}, \frac{\arcsin\left(\frac{1}{3}\right) + 2\pi}{2}, \frac{\pi - \arcsin\left(\frac{1}{3}\right) + 2\pi}{2}

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