Trigonometry Examples

$2 \cot^{2} (x) - 3 \csc (x) = 0$

Step 1

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

$2 (\csc^{2} (x) - 1) - 3 \csc (x) = 0$

Step 2

Simplify each term.

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Step 2.1

Apply the distributive property.

$2 \csc^{2} (x) + 2 \cdot - 1 - 3 \csc (x) = 0$

Step 2.2

Multiply $2$ by $- 1$ .

$2 \csc^{2} (x) - 2 - 3 \csc (x) = 0$

Step 3

Reorder the polynomial.

$2 \csc^{2} (x) - 3 \csc (x) - 2 = 0$

Step 4

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

$2 {(u)}^{2} - 3 u - 2 = 0$

Step 5

Factor by grouping.

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Step 5.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = - 3$ .

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Step 5.1.1

Factor $- 3$ out of $- 3 u$ .

$2 u^{2} - 3 u - 2 = 0$

Step 5.1.2

Rewrite $- 3$ as $1$ plus $- 4$

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

Step 5.1.3

Apply the distributive property.

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

Step 5.1.4

Multiply $u$ by $1$ .

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

Step 5.2

Factor out the greatest common factor from each group.

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Step 5.2.1

Group the first two terms and the last two terms.

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

Step 5.2.2

Factor out the greatest common factor (GCF) from each group.

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

Step 5.3

Factor the polynomial by factoring out the greatest common factor, $2 u + 1$ .

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

Step 6

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

$2 u + 1 = 0$

$u - 2 = 0$

Step 7

Set $2 u + 1$ equal to $0$ and solve for $u$ .

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Step 7.1

Set $2 u + 1$ equal to $0$ .

$2 u + 1 = 0$

Step 7.2

Solve $2 u + 1 = 0$ for $u$ .

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Step 7.2.1

Subtract $1$ from both sides of the equation.

$2 u = - 1$

Step 7.2.2

Divide each term in $2 u = - 1$ by $2$ and simplify.

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Step 7.2.2.1

Divide each term in $2 u = - 1$ by $2$ .

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 7.2.2.2

Simplify the left side.

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Step 7.2.2.2.1

Cancel the common factor of $2$ .

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Step 7.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{- 1}{2}$

Step 7.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{- 1}{2}$

Step 7.2.2.3

Simplify the right side.

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Step 7.2.2.3.1

Move the negative in front of the fraction.

$u = - \frac{1}{2}$

Step 8

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 8.1

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

$u - 2 = 0$

Step 8.2

Add $2$ to both sides of the equation.

$u = 2$

Step 9

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

$u = - \frac{1}{2}, 2$

Step 10

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

$\csc (x) = - \frac{1}{2}, 2$

Step 11

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

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

$\csc (x) = 2$

Step 12

Solve for $x$ in $\csc (x) = - \frac{1}{2}$ .

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Step 12.1

The range of cosecant is $y \leq - 1$ and $y \geq 1$ . Since $- \frac{1}{2}$ does not fall in this range, there is no solution.

No solution

Step 13

Solve for $x$ in $\csc (x) = 2$ .

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Step 13.1

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

$x = arccsc (2)$

Step 13.2

Simplify the right side.

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Step 13.2.1

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

$x = \frac{π}{6}$

Step 13.3

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.

$x = π - \frac{π}{6}$

Step 13.4

Simplify $π - \frac{π}{6}$ .

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Step 13.4.1

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 13.4.2

Combine fractions.

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Step 13.4.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 13.4.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 6 - π}{6}$

Step 13.4.3

Simplify the numerator.

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Step 13.4.3.1

Move $6$ to the left of $π$ .

$x = \frac{6 \cdot π - π}{6}$

Step 13.4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{6}$

Step 13.5

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

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Step 13.5.1

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

$\frac{2 π}{| b |}$

Step 13.5.2

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

$\frac{2 π}{| 1 |}$

Step 13.5.3

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

$\frac{2 π}{1}$

Step 13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 13.6

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

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

Step 14

List all of the solutions.

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