Precalculus Examples

$(\cot (x) + 1) (\csc (x) + 1) = 0$

Step 1

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

$\cot (x) + 1 = 0$

$\csc (x) + 1 = 0$

Step 2

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

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Step 2.1

Set $\cot (x) + 1$ equal to $0$ .

$\cot (x) + 1 = 0$

Step 2.2

Solve $\cot (x) + 1 = 0$ for $x$ .

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Step 2.2.1

Subtract $1$ from both sides of the equation.

$\cot (x) = - 1$

Step 2.2.2

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

$x = arccot (- 1)$

Step 2.2.3

Simplify the right side.

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Step 2.2.3.1

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

$x = \frac{3 π}{4}$

Step 2.2.4

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.

$x = \frac{3 π}{4} - π$

Step 2.2.5

Simplify the expression to find the second solution.

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Step 2.2.5.1

Add $2 π$ to $\frac{3 π}{4} - π$ .

$x = \frac{3 π}{4} - π + 2 π$

Step 2.2.5.2

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

$x = \frac{7 π}{4}$

Step 2.2.6

Find the period of $\cot (x)$ .

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Step 2.2.6.1

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

$\frac{π}{| b |}$

Step 2.2.6.2

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

$\frac{π}{| 1 |}$

Step 2.2.6.3

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

$\frac{π}{1}$

Step 2.2.6.4

Divide $π$ by $1$ .

$π$

Step 2.2.7

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

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

Step 3

Set $\csc (x) + 1$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\csc (x) + 1$ equal to $0$ .

$\csc (x) + 1 = 0$

Step 3.2

Solve $\csc (x) + 1 = 0$ for $x$ .

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Step 3.2.1

Subtract $1$ from both sides of the equation.

$\csc (x) = - 1$

Step 3.2.2

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

$x = arccsc (- 1)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

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

$x = - \frac{π}{2}$

Step 3.2.4

The cosecant function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 3.2.5

Simplify the expression to find the second solution.

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Step 3.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 3.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 3.2.6

Find the period of $\csc (x)$ .

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

Add $2 π$ to every negative angle to get positive angles.

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Step 3.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 3.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 3.2.7.3

Combine fractions.

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Step 3.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 3.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 3.2.7.4

Simplify the numerator.

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Step 3.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 3.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 3.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 3.2.8

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

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

Step 4

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

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

Step 5

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

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