Solve for θ in Degrees 6tan(theta)^2-10tan(theta)+1=-5tan(theta)

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Trigonometry

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

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

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

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

$6 \tan^{2} (θ) + 1 - 5 \tan (θ) = 0$

Factor by grouping.

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Reorder terms.

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

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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Factor $- 5$ out of $- 5 \tan (θ)$ .

$6 \tan^{2} (θ) - 5 \tan (θ) + 1 = 0$

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

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

Apply the distributive property.

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

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

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

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

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

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

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

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

$3 \tan (θ) - 1 = 0$

$2 \tan (θ) - 1 = 0$

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

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Set $3 \tan (θ) - 1$ equal to $0$ .

$3 \tan (θ) - 1 = 0$

Solve $3 \tan (θ) - 1 = 0$ for $θ$ .

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Add $1$ to both sides of the equation.

$3 \tan (θ) = 1$

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

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

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Simplify the left side.

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Cancel the common factor of $3$ .

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Cancel the common factor.

$\frac{3 \tan (θ)}{3} = \frac{1}{3}$

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{1}{3}$

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

$θ = \arctan (\frac{1}{3})$

Simplify the right side.

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

$θ = 18.43494882$

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.

$θ = 180 + 18.43494882$

Add $180$ and $18.43494882$ .

$θ = 198.43494882$

Find the period of $\tan (θ)$ .

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

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

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

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

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

$\frac{180}{1}$

Divide $180$ by $1$ .

$180$

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 $n$

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

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Set $2 \tan (θ) - 1$ equal to $0$ .

$2 \tan (θ) - 1 = 0$

Solve $2 \tan (θ) - 1 = 0$ for $θ$ .

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Add $1$ to both sides of the equation.

$2 \tan (θ) = 1$

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

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

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \tan (θ)}{2} = \frac{1}{2}$

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{1}{2}$

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

$θ = \arctan (\frac{1}{2})$

Simplify the right side.

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

$θ = 26.56505117$

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.

$θ = 180 + 26.56505117$

Add $180$ and $26.56505117$ .

$θ = 206.56505117$

Find the period of $\tan (θ)$ .

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

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

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

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

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

$\frac{180}{1}$

Divide $180$ by $1$ .

$180$

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 $n$

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 $n$

Consolidate the answers.

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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 $n$

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 $n$

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