Trigonometry Examples

{csc}^{2} (x) - csc (x) - 2 = 0

$\csc^{2} (x) - \csc (x) - 2 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let

$u = \csc (x)$ . Substitute

$u$ for all occurrences of

$\csc (x)$ .

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

Step 1.2

Factor

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

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Step 1.2.1

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

$- 1$ .

$- 2, 1$

Step 1.2.2

Write the factored form using these integers.

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

Step 1.3

Replace all occurrences of

$u$ with

$\csc (x)$ .

$(\csc (x) - 2) (\csc (x) + 1) = 0$

Step 2

If any individual factor on the left side of the equation is equal to

$0$ , the entire expression will be equal to

$0$ .

$\csc (x) - 2 = 0$

$\csc (x) + 1 = 0$

Step 3

Set

$\csc (x) - 2$ equal to

$0$ and solve for

$x$ .

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Step 3.1

Set

$\csc (x) - 2$ equal to

$0$ .

$\csc (x) - 2 = 0$

Step 3.2

Solve

$\csc (x) - 2 = 0$ for

$x$ .

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Step 3.2.1

Add

$2$ to both sides of the equation.

$\csc (x) = 2$

Step 3.2.2

Take the inverse cosecant of both sides of the equation to extract

$x$ from inside the cosecant.

$x = arccsc (2)$

Step 3.2.3

Simplify the right side.

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Step 3.2.3.1

The exact value of

$arccsc (2)$ is

$\frac{π}{6}$ .

$x = \frac{π}{6}$

Step 3.2.4

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.

$x = π - \frac{π}{6}$

Step 3.2.5

Simplify

$π - \frac{π}{6}$ .

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Step 3.2.5.1

To write

$π$ as a fraction with a common denominator, multiply by

$\frac{6}{6}$ .

$x = π \cdot \frac{6}{6} - \frac{π}{6}$

Step 3.2.5.2

Combine fractions.

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Step 3.2.5.2.1

Combine

$π$ and

$\frac{6}{6}$ .

$x = \frac{π \cdot 6}{6} - \frac{π}{6}$

Step 3.2.5.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 3.2.5.3

Simplify the numerator.

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Step 3.2.5.3.1

Move

$6$ to the left of

$π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 3.2.5.3.2

Subtract

$π$ from

$6 π$ .

$x = \frac{5 π}{6}$

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

The period of the

$\csc (x)$ function is

$2 π$ so values will repeat every

$2 π$ radians in both directions.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer

$n$

Step 4

Set

$\csc (x) + 1$ equal to

$0$ and solve for

$x$ .

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Step 4.1

Set

$\csc (x) + 1$ equal to

$0$ .

$\csc (x) + 1 = 0$

Step 4.2

Solve

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

$x$ .

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Step 4.2.1

Subtract

$1$ from both sides of the equation.

$\csc (x) = - 1$

Step 4.2.2

Take the inverse cosecant of both sides of the equation to extract

$x$ from inside the cosecant.

$x = arccsc (- 1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

The exact value of

$arccsc (- 1)$ is

$- \frac{π}{2}$ .

$x = - \frac{π}{2}$

Step 4.2.4

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 4.2.5

Simplify the expression to find the second solution.

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Step 4.2.5.1

Subtract

$2 π$ from

$2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 4.2.5.2

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 4.2.6

Find the period of

$\csc (x)$ .

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Step 4.2.6.1

The period of the function can be calculated using

$\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace

$b$ with

$1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

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

$0$ and

$1$ is

$1$ .

$\frac{2 π}{1}$

Step 4.2.6.4

Divide

$2 π$ by

$1$ .

$2 π$

Step 4.2.7

Add

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

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Step 4.2.7.1

Add

$2 π$ to

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

$- \frac{π}{2} + 2 π$

Step 4.2.7.2

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 4.2.7.3

Combine fractions.

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Step 4.2.7.3.1

Combine

$2 π$ and

$\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 4.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 4.2.7.4

Simplify the numerator.

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Step 4.2.7.4.1

Multiply

$2$ by

$2$ .

$\frac{4 π - π}{2}$

Step 4.2.7.4.2

Subtract

$π$ from

$4 π$ .

$\frac{3 π}{2}$

Step 4.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 4.2.8

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

$n$

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

$n$

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

$n$

Step 5

The final solution is all the values that make

$(\csc (x) - 2) (\csc (x) + 1) = 0$ true.

$x = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer

$n$

Step 6

Consolidate the answers.

$x = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer

$n$