Find The Value Of T In The Interval [0, 2n) That Satisfies The Given Equation. Tan T = 3, Csct < 0
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Introduction
In trigonometry, solving equations that involve the tangent function and other trigonometric ratios is a fundamental skill. These problems often require understanding the behavior of trigonometric functions within specific intervals and applying the properties of the unit circle.
The problem at hand asks us to find the value of T in the interval [0, 2π) that satisfies the equations:
- Tan T = 3
- Csc T < 0
This type of problem combines the solution of a basic tangent equation with an inequality involving cosecant, which is the reciprocal of sine. To approach this problem systematically, we will explore the properties of tangent and cosecant functions, analyze their signs within the interval, and determine the specific values of T that satisfy both conditions.
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Understanding the Trigonometric Functions Involved
Before diving into the solution, it is essential to understand the behavior and properties of the functions involved:
- The Tangent Function (Tan T)
- Defined as Tan T = sin T / cos T
- Periodicity: Period is π, meaning Tan T repeats every π radians.
- Domain restrictions: Tan T is undefined where cos T = 0, i.e., at T = π/2 + kπ, where k is an integer.
- Range: (-∞, +∞)
- The Cosecant Function (Csc T)
- Defined as Csc T = 1 / sin T
- Sign behavior:
- Csc T > 0 when sin T > 0
- Csc T < 0 when sin T < 0
- Domain restrictions: Csc T is undefined where sin T = 0, i.e., at T = kπ
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Step-by-Step Solution Approach
To find the value(s) of T satisfying both Tan T = 3 and Csc T < 0, follow these steps:
Step 1: Find all T where Tan T = 3 in the interval [0, 2π)
Step 2: Determine the sign of sin T at these solutions to check the inequality Csc T < 0
Step 3: Verify which solutions satisfy Csc T < 0 (i.e., sin T < 0)
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Step 1: Solving Tan T = 3 in [0, 2π)
Since Tan T = 3, T is one of the angles where the tangent value equals 3.
General solution:
- T = arctangent(3) + kπ, where k is an integer, because the tangent function repeats every π.
Calculate arctangent(3):
Using a calculator or inverse tangent function:
- arctangent(3) ≈ 1.249 radians
Find the solutions within [0, 2π):
- For k = 0:
T₁ ≈ 1.249 radians
- For k = 1:
T₂ ≈ 1.249 + π ≈ 1.249 + 3.142 ≈ 4.391 radians
- For k = 2:
T ≈ 1.249 + 2π ≈ 1.249 + 6.283 ≈ 7.532 radians
But 7.532 > 2π (~6.283), so discard.
- For k = -1:
T ≈ 1.249 - π ≈ 1.249 - 3.142 ≈ -1.893 radians, which is outside [0, 2π) (negative), so discard for the interval.
Final solutions for Tan T = 3 in [0, 2π):
- T ≈ 1.249 radians
- T ≈ 4.391 radians
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Step 2: Analyzing the Sign of sin T at these solutions
Recall:
- Csc T < 0 ⇨ sin T < 0
Now, evaluate the sine at these solutions:
For T ≈ 1.249 radians:
- Since 1.249 radians ≈ 71.6°, which is in the first quadrant where sin T > 0.
- Therefore, Csc T > 0 at this point, not satisfying Csc T < 0.
For T ≈ 4.391 radians:
- Convert to degrees:
4.391 radians ≈ 251.6°
- This angle is in the third quadrant, where sin T < 0.
- Therefore, Csc T < 0 at T ≈ 4.391 radians.
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Step 3: Final answer
The only solution within [0, 2π) where Tan T = 3 and Csc T < 0 is:
- T ≈ 4.391 radians
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Additional Considerations
Validity within the interval:
- The interval is [0, 2π), which corresponds to [0, 6.283) radians.
- Our solution T ≈ 4.391 radians lies within this interval.
Exact form:
- The exact solution for T is:
T = arctangent(3) + π ≈ 1.249 + 3.142 ≈ 4.391 radians
Note on other solutions:
- The other solution T ≈ 1.249 radians does not satisfy Csc T < 0 because sin T > 0 there.
- No other solutions in [0, 2π) satisfy both conditions.
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Summary
| Step | Description | Result |
|--------|--------------|---------|
| 1 | Find solutions of Tan T = 3 in [0, 2π) | T ≈ 1.249 radians, T ≈ 4.391 radians |
| 2 | Determine sign of sin T at these points | sin T > 0 at 1.249 rad, sin T < 0 at 4.391 rad |
| 3 | Check which satisfy Csc T < 0 | Only T ≈ 4.391 radians satisfies Csc T < 0 |
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Conclusion
The value of T in the interval [0, 2π) that satisfies the equations:
- Tan T = 3
- Csc T < 0
is approximately:
> T ≈ 4.391 radians
or in exact terms:
> T = arctangent(3) + π
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Additional Tips for Solving Similar Problems
- Always analyze the signs of the involved functions within the specified interval.
- Remember the periodicity of trigonometric functions: Tan T repeats every π, Sin T and Csc T repeat every 2π.
- Use inverse functions to find principal solutions, then adjust using periodicity to find all solutions within the interval.
- Always verify the solution by checking the sign of the sine or cosine to ensure it satisfies the inequality.
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Final Remarks
Understanding the behavior of trigonometric functions and their signs in different quadrants is crucial when solving equations involving inequalities. This problem exemplifies the importance of combining algebraic solutions with sign analysis to find the correct solutions within a given interval. Whether for academic purposes or practical applications in physics and engineering, mastering these techniques enhances problem-solving skills and deepens comprehension of trigonometric concepts.
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