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Math  /  Trigonometry

QuestionProblem 2. (1 point) Find all solutions xx to the equation 6cos⁡(π4x)=5.56 \cos \left(\frac{\pi}{4} x\right)=5.5 in the interval [0,6][0,6] (if there is thore than one solution, separate them with commas). x=□.x=\square .

Studdy Solution

STEP 1

1. The equation 6cos⁡(π4x)=5.5 6 \cos \left(\frac{\pi}{4} x\right) = 5.5 is a trigonometric equation.
2. We are looking for solutions within the interval [0,6][0, 6].
3. The cosine function is periodic, and we may have multiple solutions within the given interval.

STEP 2

1. Isolate the cosine function.
2. Solve for the angle.
3. Find all solutions for x x in the given interval.

STEP 3

First, divide both sides of the equation by 6 to isolate the cosine function:
6cos⁡(π4x)=5.5 6 \cos \left(\frac{\pi}{4} x\right) = 5.5 cos⁡(π4x)=5.56 \cos \left(\frac{\pi}{4} x\right) = \frac{5.5}{6} cos⁡(π4x)=1112 \cos \left(\frac{\pi}{4} x\right) = \frac{11}{12}

STEP 4

Now, solve for the angle π4x\frac{\pi}{4} x by taking the inverse cosine (arccos) of both sides:
π4x=cos⁡−1(1112) \frac{\pi}{4} x = \cos^{-1} \left(\frac{11}{12}\right)
Calculate the principal value of cos⁡−1(1112)\cos^{-1} \left(\frac{11}{12}\right):
Let θ=cos⁡−1(1112)\theta = \cos^{-1} \left(\frac{11}{12}\right).

STEP 5

Since the cosine function is periodic with period 2π2\pi, the general solutions for the angle are:
π4x=θ+2kπandπ4x=−θ+2kπ \frac{\pi}{4} x = \theta + 2k\pi \quad \text{and} \quad \frac{\pi}{4} x = -\theta + 2k\pi
where kk is an integer.

STEP 6

Solve for x x in both cases:
1. π4x=θ+2kπ\frac{\pi}{4} x = \theta + 2k\pi x=4π(θ+2kπ) x = \frac{4}{\pi}(\theta + 2k\pi)
2. π4x=−θ+2kπ\frac{\pi}{4} x = -\theta + 2k\pi x=4π(−θ+2kπ) x = \frac{4}{\pi}(-\theta + 2k\pi)
Calculate the specific values of x x for k k such that x x is in the interval [0,6][0, 6].

STEP 7

Calculate θ=cos⁡−1(1112)\theta = \cos^{-1} \left(\frac{11}{12}\right) using a calculator:
θ≈0.4510\theta \approx 0.4510 radians.
Substitute θ\theta back into the equations for x x :
1. x=4π(0.4510+2kπ) x = \frac{4}{\pi}(0.4510 + 2k\pi)
2. x=4π(−0.4510+2kπ) x = \frac{4}{\pi}(-0.4510 + 2k\pi)
Test integer values of k k to find solutions in the interval [0,6][0, 6].

STEP 8

For k=0 k = 0 :
1. x≈4π(0.4510)≈0.573 x \approx \frac{4}{\pi}(0.4510) \approx 0.573
2. x≈4π(−0.4510)≈−0.573 x \approx \frac{4}{\pi}(-0.4510) \approx -0.573 (not in the interval)
For k=1 k = 1 :
1. x≈4π(0.4510+2π)≈5.146 x \approx \frac{4}{\pi}(0.4510 + 2\pi) \approx 5.146
2. x≈4π(−0.4510+2π)≈4.573 x \approx \frac{4}{\pi}(-0.4510 + 2\pi) \approx 4.573
The solutions within the interval [0,6][0, 6] are:
x=0.573,4.573,5.146 x = 0.573, 4.573, 5.146
The solutions for x x in the interval [0,6][0, 6] are:
0.573,4.573,5.146 \boxed{0.573, 4.573, 5.146}

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