Solve for θ in Degrees 2cot(theta)-3=0

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Trigonometry

$2 \cot (θ) - 3 = 0$

Add $3$ to both sides of the equation.

$2 \cot (θ) = 3$

Divide each term in $2 \cot (θ) = 3$ by $2$ and simplify.

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Divide each term in $2 \cot (θ) = 3$ by $2$ .

$\frac{2 \cot (θ)}{2} = \frac{3}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \cot (θ)}{2} = \frac{3}{2}$

Divide $\cot (θ)$ by $1$ .

$\cot (θ) = \frac{3}{2}$

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

$θ = arccot (\frac{3}{2})$

Simplify the right side.

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Evaluate $arccot (\frac{3}{2})$ .

$θ = 33.69006752$

The cotangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 33.69006752$

Add $180$ and $33.69006752$ .

$θ = 213.69006752$

Find the period of $\cot (θ)$ .

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The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

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

$\frac{180}{| 1 |}$

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

$\frac{180}{1}$

Divide $180$ by $1$ .

$180$

The period of the $\cot (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 33.69006752 + 180 n, 213.69006752 + 180 n$ , for any integer $n$

Consolidate the answers.

$θ = 33.69006752 + 180 n$ , for any integer $n$

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