Solve for θ in Degrees 8tan(theta)^2+10tan(theta)+10=7

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Trigonometry

$8 \tan^{2} (θ) + 10 \tan (θ) + 10 = 7$

Subtract $7$ from both sides of the equation.

$8 \tan^{2} (θ) + 10 \tan (θ) + 10 - 7 = 0$

Subtract $7$ from $10$ .

$8 \tan^{2} (θ) + 10 \tan (θ) + 3 = 0$

Factor by grouping.

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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 $a \cdot c = 8 \cdot 3 = 24$ and whose sum is $b = 10$ .

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Factor $10$ out of $10 \tan (θ)$ .

$8 \tan^{2} (θ) + 10 (\tan (θ)) + 3 = 0$

Rewrite $10$ as $4$ plus $6$

$8 \tan^{2} (θ) + (4 + 6) \tan (θ) + 3 = 0$

Apply the distributive property.

$8 \tan^{2} (θ) + 4 \tan (θ) + 6 \tan (θ) + 3 = 0$

Factor out the greatest common factor from each group.

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Group the first two terms and the last two terms.

$(8 \tan^{2} (θ) + 4 \tan (θ)) + 6 \tan (θ) + 3 = 0$

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

$4 \tan (θ) (2 \tan (θ) + 1) + 3 (2 \tan (θ) + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 \tan (θ) + 1$ .

$(2 \tan (θ) + 1) (4 \tan (θ) + 3) = 0$

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

$2 \tan (θ) + 1 = 0$

$4 \tan (θ) + 3 = 0$

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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Subtract $1$ from 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 $1$ .

$\tan (θ) = \frac{- 1}{2}$

Simplify the right side.

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Move the negative in front of the fraction.

$\tan (θ) = - \frac{1}{2}$

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 negative in the second and fourth quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the third quadrant.

$θ = - 26.56505117 - 180$

Simplify the expression to find the second solution.

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Add $360 °$ to $- 26.56505117 - 180 °$ .

$θ = - 26.56505117 - 180 ° + 360 °$

The resulting angle of $153.43494882 °$ is positive and coterminal with $- 26.56505117 - 180$ .

$θ = 153.43494882 °$

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 $1$ .

$180$

Add $180$ to every negative angle to get positive angles.

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Add $180$ to $- 26.56505117$ to find the positive angle.

$- 26.56505117 + 180$

Subtract $26.56505117$ from $180$ .

$153.43494882$

List the new angles.

$θ = 153.43494882$

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

$θ = 153.43494882 + 180 n, 153.43494882 + 180 n$ , for any integer $n$

Set $4 \tan (θ) + 3$ equal to $0$ and solve for $θ$ .

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Set $4 \tan (θ) + 3$ equal to $0$ .

$4 \tan (θ) + 3 = 0$

Solve $4 \tan (θ) + 3 = 0$ for $θ$ .

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Subtract $3$ from both sides of the equation.

$4 \tan (θ) = - 3$

Divide each term in $4 \tan (θ) = - 3$ by $4$ and simplify.

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Divide each term in $4 \tan (θ) = - 3$ by $4$ .

$\frac{4 \tan (θ)}{4} = \frac{- 3}{4}$

Simplify the left side.

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Cancel the common factor of $4$ .

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Cancel the common factor.

$\frac{4 \tan (θ)}{4} = \frac{- 3}{4}$

Divide $\tan (θ)$ by $1$ .

$\tan (θ) = \frac{- 3}{4}$

Simplify the right side.

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Move the negative in front of the fraction.

$\tan (θ) = - \frac{3}{4}$

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

$θ = \arctan (- \frac{3}{4})$

Simplify the right side.

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

$θ = - 36.86989764$

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

$θ = - 36.86989764 - 180$

Simplify the expression to find the second solution.

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Add $360 °$ to $- 36.86989764 - 180 °$ .

$θ = - 36.86989764 - 180 ° + 360 °$

The resulting angle of $143.13010235 °$ is positive and coterminal with $- 36.86989764 - 180$ .

$θ = 143.13010235 °$

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 $1$ .

$180$

Add $180$ to every negative angle to get positive angles.

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Add $180$ to $- 36.86989764$ to find the positive angle.

$- 36.86989764 + 180$

Subtract $36.86989764$ from $180$ .

$143.13010235$

List the new angles.

$θ = 143.13010235$

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

$θ = 143.13010235 + 180 n, 143.13010235 + 180 n$ , for any integer $n$

The final solution is all the values that make $(2 \tan (θ) + 1) (4 \tan (θ) + 3) = 0$ true.

$θ = 153.43494882 + 180 n, 143.13010235 + 180 n$ , for any integer $n$

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