Find All Complex Solutions (cot(theta)+1)(csc(theta)-1)=0

$(\cot (θ) + 1) (\csc (θ) - 1) = 0$

Step 1

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$\cot (θ) + 1 = 0$

$\csc (θ) - 1 = 0$

Step 2

Set $\cot (θ) + 1$ equal to $0$ and solve for $θ$ .

Set $\cot (θ) + 1$ equal to $0$ .

$\cot (θ) + 1 = 0$

Solve $\cot (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$\cot (θ) = - 1$

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

$θ = arccot (- 1)$

Simplify the right side.

The exact value of $arccot (- 1)$ is $\frac{3 π}{4}$ .

$θ = \frac{3 π}{4}$

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

$θ = \frac{3 π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $\frac{3 π}{4} - π$ .

$θ = \frac{3 π}{4} - π + 2 π$

The resulting angle of $\frac{7 π}{4}$ is positive and coterminal with $\frac{3 π}{4} - π$ .

$θ = \frac{7 π}{4}$

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 $1$ .

$π$

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

$θ = \frac{3 π}{4} + π n, \frac{7 π}{4} + π n$ , for any integer $n$

Step 3

Set $\csc (θ) - 1$ equal to $0$ and solve for $θ$ .

Set $\csc (θ) - 1$ equal to $0$ .

$\csc (θ) - 1 = 0$

Solve $\csc (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\csc (θ) = 1$

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

$θ = arccsc (1)$

Simplify the right side.

The exact value of $arccsc (1)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

The cosecant function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $π$ to find the solution in the second quadrant.

$θ = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$θ = π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $π$ and $\frac{2}{2}$ .

$θ = \frac{π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

Find the period of $\csc (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\csc (θ)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$θ = \frac{π}{2} + 2 π n$ , for any integer $n$

Step 4

The final solution is all the values that make $(\cot (θ) + 1) (\csc (θ) - 1) = 0$ true.

$θ = \frac{3 π}{4} + π n, \frac{7 π}{4} + π n, \frac{π}{2} + 2 π n$ , for any integer $n$

Step 5

Consolidate $\frac{3 π}{4} + π n$ and $\frac{7 π}{4} + π n$ to $\frac{3 π}{4} + π n$ .

$θ = \frac{3 π}{4} + π n, \frac{π}{2} + 2 π n$ , for any integer $n$

Find All Complex Solutions (cot(theta)+1)(csc(theta)-1)=0

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