Trigonometry Examples

$\csc^{2} (θ) + \csc (θ) - 2 = 0$

Step 1

Factor the left side of the equation.

Tap for more steps...

Step 1.1

Let $u = \csc (θ)$ . Substitute $u$ for all occurrences of $\csc (θ)$ .

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

Step 1.2

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

Tap for more steps...

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 $1$ .

$- 1, 2$

Step 1.2.2

Write the factored form using these integers.

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

Step 1.3

Replace all occurrences of $u$ with $\csc (θ)$ .

$(\csc (θ) - 1) (\csc (θ) + 2) = 0$

Step 2

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

$\csc (θ) - 1 = 0$

$\csc (θ) + 2 = 0$

Step 3

Set $\csc (θ) - 1$ equal to $0$ and solve for $θ$ .

Tap for more steps...

Step 3.1

Set $\csc (θ) - 1$ equal to $0$ .

$\csc (θ) - 1 = 0$

Step 3.2

Solve $\csc (θ) - 1 = 0$ for $θ$ .

Tap for more steps...

Step 3.2.1

Add $1$ to both sides of the equation.

$\csc (θ) = 1$

Step 3.2.2

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

$θ = arccsc (1)$

Step 3.2.3

Simplify the right side.

Tap for more steps...

Step 3.2.3.1

The exact value of $arccsc (1)$ is $90$ .

$θ = 90$

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 $180$ to find the solution in the second quadrant.

$θ = 180 - 90$

Step 3.2.5

Subtract $90$ from $180$ .

$θ = 90$

Step 3.2.6

Find the period of $\csc (θ)$ .

Tap for more steps...

Step 3.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 3.2.6.2

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

$\frac{360}{| 1 |}$

Step 3.2.6.3

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

$\frac{360}{1}$

Step 3.2.6.4

Divide $360$ by $1$ .

$360$

Step 3.2.7

The period of the $\csc (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 90 + 360 n$ , for any integer $n$

Step 4

Set $\csc (θ) + 2$ equal to $0$ and solve for $θ$ .

Tap for more steps...

Step 4.1

Set $\csc (θ) + 2$ equal to $0$ .

$\csc (θ) + 2 = 0$

Step 4.2

Solve $\csc (θ) + 2 = 0$ for $θ$ .

Tap for more steps...

Step 4.2.1

Subtract $2$ from both sides of the equation.

$\csc (θ) = - 2$

Step 4.2.2

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

$θ = arccsc (- 2)$

Step 4.2.3

Simplify the right side.

Tap for more steps...

Step 4.2.3.1

The exact value of $arccsc (- 2)$ is $- 30$ .

$θ = - 30$

Step 4.2.4

The cosecant function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 30 + 180$

Step 4.2.5

Simplify the expression to find the second solution.

Tap for more steps...

Step 4.2.5.1

Subtract $360 °$ from $360 + 30 + 180 °$ .

$θ = 360 + 30 + 180 ° - 360 °$

Step 4.2.5.2

The resulting angle of $210 °$ is positive, less than $360 °$ , and coterminal with $360 + 30 + 180$ .

$θ = 210 °$

Step 4.2.6

Find the period of $\csc (θ)$ .

Tap for more steps...

Step 4.2.6.1

The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 4.2.6.2

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

$\frac{360}{| 1 |}$

Step 4.2.6.3

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

$\frac{360}{1}$

Step 4.2.6.4

Divide $360$ by $1$ .

$360$

Step 4.2.7

Add $360$ to every negative angle to get positive angles.

Tap for more steps...

Step 4.2.7.1

Add $360$ to $- 30$ to find the positive angle.

$- 30 + 360$

Step 4.2.7.2

Subtract $30$ from $360$ .

$330$

Step 4.2.7.3

List the new angles.

$θ = 330$

Step 4.2.8

The period of the $\csc (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 210 + 360 n, 330 + 360 n$ , for any integer $n$

Step 5

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

$θ = 90 + 360 n, 210 + 360 n, 330 + 360 n$ , for any integer $n$

Step 6

Consolidate the answers.

$θ = 90 + 120 n$ , for any integer $n$