Trigonometry Examples

$6 \tan^{2} (θ) - 10 \tan (θ) + 1 = - 5 \tan (θ)$

Step 1

Add $5 \tan (θ)$ to both sides of the equation.

$6 \tan^{2} (θ) - 10 \tan (θ) + 1 + 5 \tan (θ) = 0$

Step 2

Add $- 10 \tan (θ)$ and $5 \tan (θ)$ .

$6 \tan^{2} (θ) + 1 - 5 \tan (θ) = 0$

Step 3

Factor by grouping.

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Step 3.1

Reorder terms.

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

Step 3.2

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 $a \cdot c = 6 \cdot 1 = 6$ and whose sum is $b = - 5$ .

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Step 3.2.1

Factor $- 5$ out of $- 5 \tan (θ)$ .

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

Step 3.2.2

Rewrite $- 5$ as $- 2$ plus $- 3$

$6 \tan^{2} (θ) + (- 2 - 3) \tan (θ) + 1 = 0$

Step 3.2.3

Apply the distributive property.

$6 \tan^{2} (θ) - 2 \tan (θ) - 3 \tan (θ) + 1 = 0$

Step 3.3

Factor out the greatest common factor from each group.

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Step 3.3.1

Group the first two terms and the last two terms.

$6 \tan^{2} (θ) - 2 \tan (θ) - 3 \tan (θ) + 1 = 0$

Step 3.3.2

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

$2 \tan (θ) (3 \tan (θ) - 1) - (3 \tan (θ) - 1) = 0$

Step 3.4

Factor the polynomial by factoring out the greatest common factor, $3 \tan (θ) - 1$ .

$(3 \tan (θ) - 1) (2 \tan (θ) - 1) = 0$

Step 4

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

$3 \tan (θ) - 1 = 0$

$2 \tan (θ) - 1 = 0$

Step 5

Set $3 \tan (θ) - 1$ equal to $0$ and solve for $θ$ .

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Step 5.1

Set $3 \tan (θ) - 1$ equal to $0$ .

$3 \tan (θ) - 1 = 0$

Step 5.2

Solve $3 \tan (θ) - 1 = 0$ for $θ$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$3 \tan (θ) = 1$

Step 5.2.2

Divide each term in $3 \tan (θ) = 1$ by $3$ and simplify.

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Step 5.2.2.1

Divide each term in $3 \tan (θ) = 1$ by $3$ .

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $3$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Step 5.2.2.2.1.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{1}{3}$

Step 5.2.3

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

$θ = \arctan (\frac{1}{3})$

Step 5.2.4

Simplify the right side.

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Step 5.2.4.1

Evaluate $\arctan (\frac{1}{3})$ .

$θ = 18.43494882$

Step 5.2.5

The tangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 18.43494882$

Step 5.2.6

Add $180$ and $18.43494882$ .

$θ = 198.43494882$

Step 5.2.7

Find the period of $\tan (θ)$ .

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Step 5.2.7.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 5.2.7.2

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

$\frac{180}{| 1 |}$

Step 5.2.7.3

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

$\frac{180}{1}$

Step 5.2.7.4

Divide $180$ by $1$ .

$180$

Step 5.2.8

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n$ , for any integer $n$

Step 6

Set $2 \tan (θ) - 1$ equal to $0$ and solve for $θ$ .

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Step 6.1

Set $2 \tan (θ) - 1$ equal to $0$ .

$2 \tan (θ) - 1 = 0$

Step 6.2

Solve $2 \tan (θ) - 1 = 0$ for $θ$ .

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Step 6.2.1

Add $1$ to both sides of the equation.

$2 \tan (θ) = 1$

Step 6.2.2

Divide each term in $2 \tan (θ) = 1$ by $2$ and simplify.

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Step 6.2.2.1

Divide each term in $2 \tan (θ) = 1$ by $2$ .

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Cancel the common factor of $2$ .

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Step 6.2.2.2.1.1

Cancel the common factor.

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Step 6.2.2.2.1.2

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{1}{2}$

Step 6.2.3

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

$θ = \arctan (\frac{1}{2})$

Step 6.2.4

Simplify the right side.

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Step 6.2.4.1

Evaluate $\arctan (\frac{1}{2})$ .

$θ = 26.56505117$

Step 6.2.5

The tangent function is positive in the first and third quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the fourth quadrant.

$θ = 180 + 26.56505117$

Step 6.2.6

Add $180$ and $26.56505117$ .

$θ = 206.56505117$

Step 6.2.7

Find the period of $\tan (θ)$ .

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Step 6.2.7.1

The period of the function can be calculated using $\frac{180}{| b |}$ .

$\frac{180}{| b |}$

Step 6.2.7.2

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

$\frac{180}{| 1 |}$

Step 6.2.7.3

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

$\frac{180}{1}$

Step 6.2.7.4

Divide $180$ by $1$ .

$180$

Step 6.2.8

The period of the $\tan (θ)$ function is $180$ so values will repeat every $180$ degrees in both directions.

$θ = 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer $n$

Step 7

The final solution is all the values that make $(3 \tan (θ) - 1) (2 \tan (θ) - 1) = 0$ true.

$θ = 18.43494882 + 180 n, 198.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer $n$

Step 8

Consolidate the answers.

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Step 8.1

Consolidate $18.43494882 + 180 n$ and $198.43494882 + 180 n$ to $18.43494882 + 180 n$ .

$θ = 18.43494882 + 180 n, 26.56505117 + 180 n, 206.56505117 + 180 n$ , for any integer $n$

Step 8.2

Consolidate $26.56505117 + 180 n$ and $206.56505117 + 180 n$ to $26.56505117 + 180 n$ .

$θ = 18.43494882 + 180 n, 26.56505117 + 180 n$ , for any integer $n$