Trigonometry Examples

$\cot^{2} (x) - 4 \csc (x) = - 5$

Step 1

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

$(\csc^{2} (x) - 1) - 4 \csc (x) = - 5$

Step 2

Reorder the polynomial.

$\csc^{2} (x) - 4 \csc (x) - 1 = - 5$

Step 3

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

${(u)}^{2} - 4 u - 1 = - 5$

Step 4

Add $5$ to both sides of the equation.

$u^{2} - 4 u - 1 + 5 = 0$

Step 5

Add $- 1$ and $5$ .

$u^{2} - 4 u + 4 = 0$

Step 6

Factor using the perfect square rule.

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Step 6.1

Rewrite $4$ as $2^{2}$ .

$u^{2} - 4 u + 2^{2} = 0$

Step 6.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 u = 2 \cdot u \cdot 2$

Step 6.3

Rewrite the polynomial.

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

Step 6.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 2$ .

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

Step 7

Set the $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 8

Add $2$ to both sides of the equation.

$u = 2$

Step 9

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

$\csc (x) = 2$

Step 10

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

$x = arccsc (2)$

Step 11

Simplify the right side.

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Step 11.1

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

$x = \frac{π}{6}$

Step 12

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 13

Simplify $π - \frac{π}{6}$ .

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Step 13.1

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 13.2

Combine fractions.

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Step 13.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 13.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 13.3

Simplify the numerator.

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Step 13.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 13.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 14

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

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Step 14.1

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

$\frac{2 π}{| b |}$

Step 14.2

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

$\frac{2 π}{| 1 |}$

Step 14.3

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

$\frac{2 π}{1}$

Step 14.4

Divide $2 π$ by $1$ .

$2 π$

Step 15

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 $n$