Starting with 4cos(t) = 3:
cos(t) = 3/4
Taking the inverse cosine of both sides:
t = ±cos⁻¹(3/4) + 2πk, where k is an integer.
Using a calculator, we find that cos⁻¹(3/4) ≈ 0.72.
Therefore, t ≈ 0.72 + 2πk or t ≈ -0.72 + 2πk for some integer k.
Since we are looking for solutions in the interval from 0 to 2π, we can use these two equations to find the values of k that give solutions in that interval.
For t ≈ 0.72 + 2πk, we need k to be 0, 1, or -1 in order to get a solution in the desired interval. Plugging in those values of k, we get approximately:
t ≈ 0.72
t ≈ 6.05
t ≈ 3.46
For t ≈ -0.72 + 2πk, we need k to be 0, 1, or -1 in order to get a solution in the desired interval. Plugging in those values of k, we get approximately:
t ≈ 5.41
t ≈ 2.83
t ≈ 0.25
Rounding all of these answers to the nearest hundredth, we get:
t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, 6.05
Therefore, the solutions in the interval from 0 to 2π are t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, and 6.05.
solve an equation in the interval from 0 to 2π. Round to the nearest hundredth. 4cos(t)=3
5 answers
no words just the steps
1. Start with the equation: 4cos(t) = 3
2. Divide both sides by 4: cos(t) = 3/4
3. Take the inverse cosine of both sides: t = ±cos⁻¹(3/4) + 2πk
4. Convert cos⁻¹(3/4) to a decimal using a calculator: cos⁻¹(3/4) ≈ 0.72
5. Plug in values of k to get solutions in the interval from 0 to 2π:
- for t ≈ 0.72 + 2πk: k = 0, 1, -1. t ≈ 0.72, 6.05, 3.46
- for t ≈ -0.72 + 2πk: k = 0, 1, -1. t ≈ 5.41, 2.83, 0.25
6. Round the solutions to the nearest hundredth: t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, 6.05
7. Therefore, the solutions in the interval from 0 to 2π are t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, and 6.05.
2. Divide both sides by 4: cos(t) = 3/4
3. Take the inverse cosine of both sides: t = ±cos⁻¹(3/4) + 2πk
4. Convert cos⁻¹(3/4) to a decimal using a calculator: cos⁻¹(3/4) ≈ 0.72
5. Plug in values of k to get solutions in the interval from 0 to 2π:
- for t ≈ 0.72 + 2πk: k = 0, 1, -1. t ≈ 0.72, 6.05, 3.46
- for t ≈ -0.72 + 2πk: k = 0, 1, -1. t ≈ 5.41, 2.83, 0.25
6. Round the solutions to the nearest hundredth: t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, 6.05
7. Therefore, the solutions in the interval from 0 to 2π are t ≈ 0.72, 0.25, 2.83, 3.46, 5.41, and 6.05.
sorry, but
cos(t) = 3/4
has only two solutions, in QI and QIV
cos(t) = 3/4
has only two solutions, in QI and QIV
You are correct. I apologize for the mistake in my previous response.
Here are the corrected steps:
1. Start with the equation: 4cos(t) = 3
2. Divide both sides by 4: cos(t) = 3/4
3. Take the inverse cosine of both sides: t = ±cos⁻¹(3/4) + 2πk
4. Convert cos⁻¹(3/4) to a decimal using a calculator: cos⁻¹(3/4) ≈ 0.72
5. Since cosine is positive in Quadrant I and Quadrant IV, we only need to consider those two quadrants.
6. Plug in values of k to get solutions in the interval from 0 to 2π:
- for t ≈ 0.72 + 2πk: k = 0. t ≈ 0.72
- for t ≈ -0.72 + 2πk: k = 0, t ≈ 2π - 0.72
7. Round the solutions to the nearest hundredth: t ≈ 0.72, 5.56
8. Therefore, the solutions in the interval from 0 to 2π are t ≈ 0.72 and t ≈ 5.56.
Here are the corrected steps:
1. Start with the equation: 4cos(t) = 3
2. Divide both sides by 4: cos(t) = 3/4
3. Take the inverse cosine of both sides: t = ±cos⁻¹(3/4) + 2πk
4. Convert cos⁻¹(3/4) to a decimal using a calculator: cos⁻¹(3/4) ≈ 0.72
5. Since cosine is positive in Quadrant I and Quadrant IV, we only need to consider those two quadrants.
6. Plug in values of k to get solutions in the interval from 0 to 2π:
- for t ≈ 0.72 + 2πk: k = 0. t ≈ 0.72
- for t ≈ -0.72 + 2πk: k = 0, t ≈ 2π - 0.72
7. Round the solutions to the nearest hundredth: t ≈ 0.72, 5.56
8. Therefore, the solutions in the interval from 0 to 2π are t ≈ 0.72 and t ≈ 5.56.
7 answers