Evaluate square root of 3cot(theta)-1=0

$\sqrt{3} \cot (θ) - 1 = 0$

Step 1

Add $1$ to both sides of the equation.

$\sqrt{3} \cot (θ) = 1$

Step 2

Divide each term in $\sqrt{3} \cot (θ) = 1$ by $\sqrt{3}$ and simplify.

Divide each term in $\sqrt{3} \cot (θ) = 1$ by $\sqrt{3}$ .

$\frac{\sqrt{3} \cot (θ)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$

Simplify the left side.

Cancel the common factor of $\sqrt{3}$ .

Cancel the common factor.

$\frac{\sqrt{3} \cot (θ)}{\sqrt{3}} = \frac{1}{\sqrt{3}}$

Divide $\cot (θ)$ by $1$ .

$\cot (θ) = \frac{1}{\sqrt{3}}$

Simplify the right side.

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\cot (θ) = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}$

Combine and simplify the denominator.

Multiply $\frac{1}{\sqrt{3}}$ by $\frac{\sqrt{3}}{\sqrt{3}}$ .

$\cot (θ) = \frac{\sqrt{3}}{\sqrt{3} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\cot (θ) = \frac{\sqrt{3}}{{\sqrt{3}}^{1} \sqrt{3}}$

Raise $\sqrt{3}$ to the power of $1$ .

$\cot (θ) = \frac{\sqrt{3}}{{\sqrt{3}}^{1} {\sqrt{3}}^{1}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\cot (θ) = \frac{\sqrt{3}}{{\sqrt{3}}^{1 + 1}}$

Add $1$ and $1$ .

$\cot (θ) = \frac{\sqrt{3}}{{\sqrt{3}}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\cot (θ) = \frac{\sqrt{3}}{{(3^{\frac{1}{2}})}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\cot (θ) = \frac{\sqrt{3}}{3^{\frac{1}{2} \cdot 2}}$

Combine $\frac{1}{2}$ and $2$ .

$\cot (θ) = \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Cancel the common factor of $2$ .

Cancel the common factor.

$\cot (θ) = \frac{\sqrt{3}}{3^{\frac{2}{2}}}$

Rewrite the expression.

$\cot (θ) = \frac{\sqrt{3}}{3^{1}}$

Evaluate the exponent.

$\cot (θ) = \frac{\sqrt{3}}{3}$

Step 3

Take the inverse cotangent of both sides of the equation to extract $θ$ from inside the cotangent.

$θ = arccot (\frac{\sqrt{3}}{3})$

Step 4

Simplify the right side.

The exact value of $arccot (\frac{\sqrt{3}}{3})$ is $\frac{π}{3}$ .

$θ = \frac{π}{3}$

Step 5

The cotangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + \frac{π}{3}$

Step 6

Simplify $π + \frac{π}{3}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = π \cdot \frac{3}{3} + \frac{π}{3}$

Combine fractions.

Combine $π$ and $\frac{3}{3}$ .

$θ = \frac{π \cdot 3}{3} + \frac{π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 3 + π}{3}$

Simplify the numerator.

Move $3$ to the left of $π$ .

$θ = \frac{3 \cdot π + π}{3}$

Add $3 π$ and $π$ .

$θ = \frac{4 π}{3}$

Step 7

Find the period of $\cot (θ)$ .

The period of the function can be calculated using $\frac{π}{| b |}$ .

$\frac{π}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{π}{| 1 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

Step 8

The period of the $\cot (θ)$ function is $π$ so values will repeat every $π$ radians in both directions.

$θ = \frac{π}{3} + π n, \frac{4 π}{3} + π n$ , for any integer $n$

Step 9

Consolidate the answers.

$θ = \frac{π}{3} + π n$ , for any integer $n$

Evaluate square root of 3cot(theta)-1=0

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