Algebra Examples

$1 = \cot^{2} (x) + \csc (x)$

Step 1

Rewrite the equation as $\cot^{2} (x) + \csc (x) = 1$ .

$\cot^{2} (x) + \csc (x) = 1$

Step 2

Replace the $\cot^{2} (x)$ with $\csc^{2} (x) - 1$ based on the $\cot^{2} (x) + 1 = \csc^{2} (x)$ identity.

$(\csc^{2} (x) - 1) + \csc (x) = 1$

Step 3

Reorder the polynomial.

$\csc^{2} (x) + \csc (x) - 1 = 1$

Step 4

Substitute $u$ for $\csc (x)$ .

${(u)}^{2} + u - 1 = 1$

Step 5

Subtract $1$ from both sides of the equation.

$u^{2} + u - 1 - 1 = 0$

Step 6

Subtract $1$ from $- 1$ .

$u^{2} + u - 2 = 0$

Step 7

Factor $u^{2} + u - 2$ using the AC method.

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Step 7.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $1$ .

$- 1, 2$

Step 7.2

Write the factored form using these integers.

$(u - 1) (u + 2) = 0$

Step 8

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

$u - 1 = 0$

$u + 2 = 0$

Step 9

Set $u - 1$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 9.2

Add $1$ to both sides of the equation.

$u = 1$

Step 10

Set $u + 2$ equal to $0$ and solve for $u$ .

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Step 10.1

Set $u + 2$ equal to $0$ .

$u + 2 = 0$

Step 10.2

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 11

The final solution is all the values that make $(u - 1) (u + 2) = 0$ true.

$u = 1, - 2$

Step 12

Substitute $\csc (x)$ for $u$ .

$\csc (x) = 1, - 2$

Step 13

Set up each of the solutions to solve for $x$ .

$\csc (x) = 1$

$\csc (x) = - 2$

Step 14

Solve for $x$ in $\csc (x) = 1$ .

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Step 14.1

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

$x = arccsc (1)$

Step 14.2

Simplify the right side.

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Step 14.2.1

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

$x = \frac{π}{2}$

Step 14.3

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.

$x = π - \frac{π}{2}$

Step 14.4

Simplify $π - \frac{π}{2}$ .

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Step 14.4.1

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

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

Step 14.4.2

Combine fractions.

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Step 14.4.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 14.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 14.4.3

Simplify the numerator.

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Step 14.4.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 14.4.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 14.5

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

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Step 14.5.1

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

$\frac{2 π}{| b |}$

Step 14.5.2

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

$\frac{2 π}{| 1 |}$

Step 14.5.3

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

$\frac{2 π}{1}$

Step 14.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 14.6

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

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

Step 15

Solve for $x$ in $\csc (x) = - 2$ .

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Step 15.1

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

$x = arccsc (- 2)$

Step 15.2

Simplify the right side.

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Step 15.2.1

The exact value of $arccsc (- 2)$ is $- \frac{π}{6}$ .

$x = - \frac{π}{6}$

Step 15.3

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.

$x = 2 π + \frac{π}{6} + π$

Step 15.4

Simplify the expression to find the second solution.

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Step 15.4.1

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$x = 2 π + \frac{π}{6} + π - 2 π$

Step 15.4.2

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$x = \frac{7 π}{6}$

Step 15.5

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

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Step 15.5.1

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

$\frac{2 π}{| b |}$

Step 15.5.2

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

$\frac{2 π}{| 1 |}$

Step 15.5.3

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

$\frac{2 π}{1}$

Step 15.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 15.6

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

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Step 15.6.1

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

Step 15.6.2

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Step 15.6.3

Combine fractions.

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Step 15.6.3.1

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Step 15.6.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Step 15.6.4

Simplify the numerator.

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Step 15.6.4.1

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Step 15.6.4.2

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

Step 15.6.5

List the new angles.

$x = \frac{11 π}{6}$

Step 15.7

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

$x = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 16

List all of the solutions.

$x = \frac{π}{2} + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 17

Consolidate the answers.

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