Solve for θ tan(2theta)=-1

$\tan (2 θ) = - 1$

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

$2 θ = \arctan (- 1)$

Simplify the right side.

The exact value of $\arctan (- 1)$ is $- \frac{π}{4}$ .

$2 θ = - \frac{π}{4}$

Divide each term in $2 θ = - \frac{π}{4}$ by $2$ and simplify.

Divide each term in $2 θ = - \frac{π}{4}$ by $2$ .

$\frac{2 θ}{2} = \frac{- \frac{π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{- \frac{π}{4}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{- \frac{π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = - \frac{π}{4} \cdot \frac{1}{2}$

Multiply $- \frac{π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{1}{2}$ by $\frac{π}{4}$ .

$θ = - \frac{π}{2 \cdot 4}$

Multiply $2$ by $4$ .

$θ = - \frac{π}{8}$

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.

$2 θ = - \frac{π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{4} - π$ .

$2 θ = - \frac{π}{4} - π + 2 π$

The resulting angle of $\frac{3 π}{4}$ is positive and coterminal with $- \frac{π}{4} - π$ .

$2 θ = \frac{3 π}{4}$

Divide each term in $2 θ = \frac{3 π}{4}$ by $2$ and simplify.

Divide each term in $2 θ = \frac{3 π}{4}$ by $2$ .

$\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 θ}{2} = \frac{\frac{3 π}{4}}{2}$

Divide $θ$ by $1$ .

$θ = \frac{\frac{3 π}{4}}{2}$

Simplify the right side.

Multiply the numerator by the reciprocal of the denominator.

$θ = \frac{3 π}{4} \cdot \frac{1}{2}$

Multiply $\frac{3 π}{4} \cdot \frac{1}{2}$ .

Multiply $\frac{3 π}{4}$ by $\frac{1}{2}$ .

$θ = \frac{3 π}{4 \cdot 2}$

Multiply $4$ by $2$ .

$θ = \frac{3 π}{8}$

Find the period of $\tan (2 θ)$ .

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

$\frac{π}{| b |}$

Replace $b$ with $2$ in the formula for period.

$\frac{π}{| 2 |}$

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Add $\frac{π}{2}$ to every negative angle to get positive angles.

Add $\frac{π}{2}$ to $- \frac{π}{8}$ to find the positive angle.

$- \frac{π}{8} + \frac{π}{2}$

To write $\frac{π}{2}$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$\frac{π}{2} \cdot \frac{4}{4} - \frac{π}{8}$

Write each expression with a common denominator of $8$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{π}{2}$ by $\frac{4}{4}$ .

$\frac{π \cdot 4}{2 \cdot 4} - \frac{π}{8}$

Multiply $2$ by $4$ .

$\frac{π \cdot 4}{8} - \frac{π}{8}$

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{8}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{8}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{8}$

List the new angles.

$θ = \frac{3 π}{8}$

The period of the $\tan (2 θ)$ function is $\frac{π}{2}$ so values will repeat every $\frac{π}{2}$ radians in both directions.

$θ = \frac{3 π}{8} + \frac{π n}{2}, \frac{3 π}{8} + \frac{π n}{2}$ , for any integer $n$

Consolidate the answers.

$θ = \frac{3 π}{8} + \frac{π n}{2}$ , for any integer $n$

Solve for θ tan(2theta)=-1

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