Math Problem Statement

Solve the trigonometric equation for all values 0, is less than or equal to, x, is less than, 2, pi0≤x<2π.

square root of, 3, end square root, cotangent, x, minus, 1, equals, 0

3​cotx−1=0

Solution

We are given the trigonometric equation:

3cot(x)1=03 \cot(x) - 1 = 0

Step 1: Isolate cot(x)\cot(x)

Start by solving for cot(x)\cot(x) in the equation:

3cot(x)=13 \cot(x) = 1

Divide both sides by 3:

cot(x)=13\cot(x) = \frac{1}{3}

Step 2: Use the cotangent identity

Recall that cot(x)\cot(x) is the reciprocal of tan(x)\tan(x), so:

cot(x)=1tan(x)\cot(x) = \frac{1}{\tan(x)}

Thus, the equation becomes:

1tan(x)=13\frac{1}{\tan(x)} = \frac{1}{3}

Step 3: Solve for tan(x)\tan(x)

Taking the reciprocal of both sides:

tan(x)=3\tan(x) = 3

Step 4: Find the general solution for tan(x)=3\tan(x) = 3

The general solution for tan(x)=a\tan(x) = a is given by:

x=tan1(a)+nπwherenZx = \tan^{-1}(a) + n\pi \quad \text{where} \quad n \in \mathbb{Z}

For tan(x)=3\tan(x) = 3, we first find the principal value of xx:

x=tan1(3)x = \tan^{-1}(3)

Using a calculator or reference table:

x1.249 radiansx \approx 1.249 \text{ radians}

Step 5: Find the specific solutions within the interval 0x<2π0 \leq x < 2\pi

Since the tangent function has a period of π\pi, the solutions are:

x=1.249+nπx = 1.249 + n\pi

Now, consider the values of nn that give solutions within the range 0x<2π0 \leq x < 2\pi.

  • For n=0n = 0, x=1.249x = 1.249
  • For n=1n = 1, x=1.249+π1.249+3.1416=4.3916x = 1.249 + \pi \approx 1.249 + 3.1416 = 4.3916

Thus, the solutions within the interval 0x<2π0 \leq x < 2\pi are:

x1.249radiansandx4.392radiansx \approx 1.249 \, \text{radians} \quad \text{and} \quad x \approx 4.392 \, \text{radians}

Final Answer:

The solutions to the equation 3cot(x)1=03 \cot(x) - 1 = 0 in the interval 0x<2π0 \leq x < 2\pi are:

x1.249radiansandx4.392radiansx \approx 1.249 \, \text{radians} \quad \text{and} \quad x \approx 4.392 \, \text{radians}


Would you like further clarification on any part of the solution? Or do you have any other questions?

Related Questions:

  1. How do you find solutions for other trigonometric functions like sine and cosine?
  2. What is the relationship between tangent and cotangent functions?
  3. How can you use the unit circle to find trigonometric solutions?
  4. What is the effect of changing the interval for solving trigonometric equations?
  5. How do you solve a trigonometric equation when multiple functions are involved?

Tip: When solving trigonometric equations, always remember the periodicity of the functions (e.g., sine, cosine, tangent, and cotangent) to find all possible solutions within a given interval.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Cotangent
Tangent
Reciprocal Functions
Solving Trigonometric Equations

Formulas

cot(x) = 1/tan(x)
General solution for tan(x) = a: x = tan^-1(a) + nπ, where n ∈ ℤ

Theorems

Periodicity of the tangent function

Suitable Grade Level

Grades 10-12