Trigonometry Examples

$\sec^{2} (x) + 2 \tan (x) = 6$

Step 1

Replace the $\sec^{2} (x)$ with $1 + \tan^{2} (x)$ based on the $\tan^{2} (x) + 1 = \sec^{2} (x)$ identity.

$(1 + \tan^{2} (x)) + 2 \tan (x) = 6$

Step 2

Reorder the polynomial.

$\tan^{2} (x) + 2 \tan (x) + 1 = 6$

Step 3

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

${(u)}^{2} + 2 (u) + 1 = 6$

Step 4

Move all terms to the left side of the equation and simplify.

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Step 4.1

Subtract $6$ from both sides of the equation.

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

Step 4.2

Subtract $6$ from $1$ .

$u^{2} + 2 u - 5 = 0$

Step 5

Use the quadratic formula to find the solutions.

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

Step 6

Substitute the values $a = 1$ , $b = 2$ , and $c = - 5$ into the quadratic formula and solve for $u$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot - 5)}}{2 \cdot 1}$

Step 7

Simplify.

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Step 7.1

Simplify the numerator.

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Step 7.1.1

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

$u = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot - 5}}{2 \cdot 1}$

Step 7.1.2

Multiply $- 4 \cdot 1 \cdot - 5$ .

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Step 7.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{- 2 \pm \sqrt{4 - 4 \cdot - 5}}{2 \cdot 1}$

Step 7.1.2.2

Multiply $- 4$ by $- 5$ .

$u = \frac{- 2 \pm \sqrt{4 + 20}}{2 \cdot 1}$

Step 7.1.3

Add $4$ and $20$ .

$u = \frac{- 2 \pm \sqrt{24}}{2 \cdot 1}$

Step 7.1.4

Rewrite $24$ as $2^{2} \cdot 6$ .

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Step 7.1.4.1

Factor $4$ out of $24$ .

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

Step 7.1.4.2

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

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

Step 7.1.5

Pull terms out from under the radical.

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

Step 7.2

Multiply $2$ by $1$ .

$u = \frac{- 2 \pm 2 \sqrt{6}}{2}$

Step 7.3

Simplify $\frac{- 2 \pm 2 \sqrt{6}}{2}$ .

$u = - 1 \pm \sqrt{6}$

Step 8

The final answer is the combination of both solutions.

$u = - 1 + \sqrt{6}, - 1 - \sqrt{6}$

Step 9

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

$\tan (x) = - 1 + \sqrt{6}, - 1 - \sqrt{6}$

Step 10

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

$\tan (x) = - 1 + \sqrt{6}$

$\tan (x) = - 1 - \sqrt{6}$

Step 11

Solve for $x$ in $\tan (x) = - 1 + \sqrt{6}$ .

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Step 11.1

Convert the right side of the equation to its decimal equivalent.

$\tan (x) = 1.44948974$

Step 11.2

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

$x = \arctan (1.44948974)$

Step 11.3

Simplify the right side.

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Step 11.3.1

Evaluate $\arctan (1.44948974)$ .

$x = 0.96688248$

Step 11.4

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.

$x = (3.14159265) + 0.96688248$

Step 11.5

Solve for $x$ .

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Step 11.5.1

Remove parentheses.

$x = 3.14159265 + 0.96688248$

Step 11.5.2

Remove parentheses.

$x = (3.14159265) + 0.96688248$

Step 11.5.3

Add $3.14159265$ and $0.96688248$ .

$x = 4.10847514$

Step 11.6

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

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Step 11.6.1

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

$\frac{π}{| b |}$

Step 11.6.2

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

$\frac{π}{| 1 |}$

Step 11.6.3

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

$\frac{π}{1}$

Step 11.6.4

Divide $π$ by $1$ .

$π$

Step 11.7

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

$x = 0.96688248 + π n, 4.10847514 + π n$ , for any integer $n$

Step 12

Solve for $x$ in $\tan (x) = - 1 - \sqrt{6}$ .

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Step 12.1

Convert the right side of the equation to its decimal equivalent.

$\tan (x) = - 3.44948974$

Step 12.2

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

$x = \arctan (- 3.44948974)$

Step 12.3

Simplify the right side.

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Step 12.3.1

Evaluate $\arctan (- 3.44948974)$ .

$x = - 1.28863304$

Step 12.4

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

$x = - 1.28863304 - (3.14159265)$

Step 12.5

Simplify the expression to find the second solution.

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Step 12.5.1

Add $2 π$ to $- 1.28863304 - (3.14159265)$ .

$x = - 1.28863304 - (3.14159265) + 2 π$

Step 12.5.2

The resulting angle of $1.85295961$ is positive and coterminal with $- 1.28863304 - (3.14159265)$ .

$x = 1.85295961$

Step 12.6

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

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Step 12.6.1

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

$\frac{π}{| b |}$

Step 12.6.2

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

$\frac{π}{| 1 |}$

Step 12.6.3

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

$\frac{π}{1}$

Step 12.6.4

Divide $π$ by $1$ .

$π$

Step 12.7

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

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Step 12.7.1

Add $π$ to $- 1.28863304$ to find the positive angle.

$- 1.28863304 + π$

Step 12.7.2

Replace with decimal approximation.

$3.14159265 - 1.28863304$

Step 12.7.3

Subtract $1.28863304$ from $3.14159265$ .

$1.85295961$

Step 12.7.4

List the new angles.

$x = 1.85295961$

Step 12.8

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

$x = 1.85295961 + π n, 1.85295961 + π n$ , for any integer $n$

Step 13

List all of the solutions.

$x = 0.96688248 + π n, 4.10847514 + π n, 1.85295961 + π n, 1.85295961 + π n$ , for any integer $n$

Step 14

Consolidate the solutions.

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Step 14.1

Consolidate $0.96688248 + π n$ and $4.10847514 + π n$ to $0.96688248 + π n$ .

$x = 0.96688248 + π n, 1.85295961 + π n, 1.85295961 + π n$ , for any integer $n$

Step 14.2

Consolidate $1.85295961 + π n$ and $1.85295961 + π n$ to $1.85295961 + π n$ .

$x = 0.96688248 + π n, 1.85295961 + π n$ , for any integer $n$