Trigonometry Examples

$\csc (x) = \frac{\sqrt{28}}{5}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $28$ as $2^{2} \cdot 7$ .

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Step 1.1.1

Factor $4$ out of $28$ .

$\csc (x) = \frac{\sqrt{4 (7)}}{5}$

Step 1.1.2

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

$\csc (x) = \frac{\sqrt{2^{2} \cdot 7}}{5}$

Step 1.2

Pull terms out from under the radical.

$\csc (x) = \frac{2 \sqrt{7}}{5}$

Step 2

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

$x = arccsc (\frac{2 \sqrt{7}}{5})$

Step 3

Simplify the right side.

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Step 3.1

Evaluate $arccsc (\frac{2 \sqrt{7}}{5})$ .

$x = 1.23732315$

Step 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.

$x = (3.14159265) - 1.23732315$

Step 5

Solve for $x$ .

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Step 5.1

Remove parentheses.

$x = 3.14159265 - 1.23732315$

Step 5.2

Remove parentheses.

$x = (3.14159265) - 1.23732315$

Step 5.3

Subtract $1.23732315$ from $3.14159265$ .

$x = 1.90426949$

Step 6

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

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Step 6.1

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

$\frac{2 π}{| b |}$

Step 6.2

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

$\frac{2 π}{| 1 |}$

Step 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 6.4

Divide $2 π$ by $1$ .

$2 π$

Step 7

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

$x = 1.23732315 + 2 π n, 1.90426949 + 2 π n$ , for any integer $n$