Trigonometry Examples

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

Step 1

Since $x$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

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

Step 2

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

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

Step 3

Combine the opposite terms in $\csc^{2} (x) - 1 + 1$ .

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Step 3.1

Add $- 1$ and $1$ .

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

Step 3.2

Add $\csc^{2} (x)$ and $0$ .

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

Step 4

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

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

Step 5

Subtract $u$ from both sides of the equation.

$u^{2} - u = 0$

Step 6

Factor $u$ out of $u^{2} - u$ .

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Step 6.1

Factor $u$ out of $u^{2}$ .

$u \cdot u - u = 0$

Step 6.2

Factor $u$ out of $- u$ .

$u \cdot u + u \cdot - 1 = 0$

Step 6.3

Factor $u$ out of $u \cdot u + u \cdot - 1$ .

$u (u - 1) = 0$

Step 7

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

$u = 0$

$u - 1 = 0$

Step 8

Set $u$ equal to $0$ .

$u = 0$

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

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

$u = 0, 1$

Step 11

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

$\csc (x) = 0, 1$

Step 12

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

$\csc (x) = 0$

$\csc (x) = 1$

Step 13

Solve for $x$ in $\csc (x) = 0$ .

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Step 13.1

The range of cosecant is $y \leq - 1$ and $y \geq 1$ . Since $0$ does not fall in this range, there is no solution.

No solution

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

List all of the solutions.

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