Trigonometry Examples

$2 \tan^{2} (x) - 3 \tan (x) = - \frac{1}{2}$

Step 1

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

$2 {(u)}^{2} - 3 u = - \frac{1}{2}$

Step 2

Add $\frac{1}{2}$ to both sides of the equation.

$2 u^{2} - 3 u + \frac{1}{2} = 0$

Step 3

Multiply through by the least common denominator $2$ , then simplify.

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Step 3.1

Apply the distributive property.

$2 (2 u^{2}) + 2 (- 3 u) + 2 (\frac{1}{2}) = 0$

Step 3.2

Simplify.

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Step 3.2.1

Multiply $2$ by $2$ .

$4 u^{2} + 2 (- 3 u) + 2 (\frac{1}{2}) = 0$

Step 3.2.2

Multiply $- 3$ by $2$ .

$4 u^{2} - 6 u + 2 (\frac{1}{2}) = 0$

Step 3.2.3

Cancel the common factor of $2$ .

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Step 3.2.3.1

Cancel the common factor.

$4 u^{2} - 6 u + 2 (\frac{1}{2}) = 0$

Step 3.2.3.2

Rewrite the expression.

$4 u^{2} - 6 u + 1 = 0$

Step 4

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5

Substitute the values $a = 4$ , $b = - 6$ , and $c = 1$ into the quadratic formula and solve for $u$ .

$\frac{6 \pm \sqrt{{(- 6)}^{2} - 4 \cdot (4 \cdot 1)}}{2 \cdot 4}$

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Raise $- 6$ to the power of $2$ .

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 4 \cdot 1}}{2 \cdot 4}$

Step 6.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 6.1.2.1

Multiply $- 4$ by $4$ .

$u = \frac{6 \pm \sqrt{36 - 16 \cdot 1}}{2 \cdot 4}$

Step 6.1.2.2

Multiply $- 16$ by $1$ .

$u = \frac{6 \pm \sqrt{36 - 16}}{2 \cdot 4}$

Step 6.1.3

Subtract $16$ from $36$ .

$u = \frac{6 \pm \sqrt{20}}{2 \cdot 4}$

Step 6.1.4

Rewrite $20$ as $2^{2} \cdot 5$ .

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Step 6.1.4.1

Factor $4$ out of $20$ .

$u = \frac{6 \pm \sqrt{4 (5)}}{2 \cdot 4}$

Step 6.1.4.2

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

$u = \frac{6 \pm \sqrt{2^{2} \cdot 5}}{2 \cdot 4}$

Step 6.1.5

Pull terms out from under the radical.

$u = \frac{6 \pm 2 \sqrt{5}}{2 \cdot 4}$

Step 6.2

Multiply $2$ by $4$ .

$u = \frac{6 \pm 2 \sqrt{5}}{8}$

Step 6.3

Simplify $\frac{6 \pm 2 \sqrt{5}}{8}$ .

$u = \frac{3 \pm \sqrt{5}}{4}$

Step 7

The final answer is the combination of both solutions.

$u = \frac{3 + \sqrt{5}}{4}, \frac{3 - \sqrt{5}}{4}$

Step 8

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

$\tan (x) = \frac{3 + \sqrt{5}}{4}, \frac{3 - \sqrt{5}}{4}$

Step 9

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

$\tan (x) = \frac{3 + \sqrt{5}}{4}$

$\tan (x) = \frac{3 - \sqrt{5}}{4}$

Step 10

Solve for $x$ in $\tan (x) = \frac{3 + \sqrt{5}}{4}$ .

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Step 10.1

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

$x = \arctan (\frac{3 + \sqrt{5}}{4})$

Step 10.2

Simplify the right side.

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Step 10.2.1

Evaluate $\arctan (\frac{3 + \sqrt{5}}{4})$ .

$x = 0.91843818$

Step 10.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = (3.14159265) + 0.91843818$

Step 10.4

Solve for $x$ .

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Step 10.4.1

Remove parentheses.

$x = 3.14159265 + 0.91843818$

Step 10.4.2

Remove parentheses.

$x = (3.14159265) + 0.91843818$

Step 10.4.3

Add $3.14159265$ and $0.91843818$ .

$x = 4.06003084$

Step 10.5

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

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Step 10.5.1

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

$\frac{π}{| b |}$

Step 10.5.2

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

$\frac{π}{| 1 |}$

Step 10.5.3

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

$\frac{π}{1}$

Step 10.5.4

Divide $π$ by $1$ .

$π$

Step 10.6

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

$x = 0.91843818 + π n, 4.06003084 + π n$ , for any integer $n$

Step 11

Solve for $x$ in $\tan (x) = \frac{3 - \sqrt{5}}{4}$ .

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Step 11.1

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

$x = \arctan (\frac{3 - \sqrt{5}}{4})$

Step 11.2

Simplify the right side.

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Step 11.2.1

Evaluate $\arctan (\frac{3 - \sqrt{5}}{4})$ .

$x = 0.18871053$

Step 11.3

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x = (3.14159265) + 0.18871053$

Step 11.4

Solve for $x$ .

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Step 11.4.1

Remove parentheses.

$x = 3.14159265 + 0.18871053$

Step 11.4.2

Remove parentheses.

$x = (3.14159265) + 0.18871053$

Step 11.4.3

Add $3.14159265$ and $0.18871053$ .

$x = 3.33030318$

Step 11.5

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

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Step 11.5.1

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

$\frac{π}{| b |}$

Step 11.5.2

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

$\frac{π}{| 1 |}$

Step 11.5.3

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

$\frac{π}{1}$

Step 11.5.4

Divide $π$ by $1$ .

$π$

Step 11.6

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

$x = 0.18871053 + π n, 3.33030318 + π n$ , for any integer $n$

Step 12

List all of the solutions.

$x = 0.91843818 + π n, 4.06003084 + π n, 0.18871053 + π n, 3.33030318 + π n$ , for any integer $n$

Step 13

Consolidate the solutions.

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Step 13.1

Consolidate $0.91843818 + π n$ and $4.06003084 + π n$ to $0.91843818 + π n$ .

$x = 0.91843818 + π n, 0.18871053 + π n, 3.33030318 + π n$ , for any integer $n$

Step 13.2

Consolidate $0.18871053 + π n$ and $3.33030318 + π n$ to $0.18871053 + π n$ .

$x = 0.91843818 + π n, 0.18871053 + π n$ , for any integer $n$