Solve for θ tan(theta)=-2sin(theta)

$\tan (θ) = - 2 \sin (θ)$

Step 1

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

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

$\frac{\tan (θ)}{\tan (θ)} = \frac{- 2 \sin (θ)}{\tan (θ)}$

Simplify the left side.

Cancel the common factor of $\tan (θ)$ .

Cancel the common factor.

$\frac{\tan (θ)}{\tan (θ)} = \frac{- 2 \sin (θ)}{\tan (θ)}$

Rewrite the expression.

$1 = \frac{- 2 \sin (θ)}{\tan (θ)}$

Simplify the right side.

Separate fractions.

$1 = \frac{- 2}{1} \cdot \frac{\sin (θ)}{\tan (θ)}$

Rewrite $\tan (θ)$ in terms of sines and cosines.

$1 = \frac{- 2}{1} \cdot \frac{\sin (θ)}{\frac{\sin (θ)}{\cos (θ)}}$

Multiply by the reciprocal of the fraction to divide by $\frac{\sin (θ)}{\cos (θ)}$ .

$1 = \frac{- 2}{1} (\sin (θ) \frac{\cos (θ)}{\sin (θ)})$

Write $\sin (θ)$ as a fraction with denominator $1$ .

$1 = \frac{- 2}{1} (\frac{\sin (θ)}{1} \cdot \frac{\cos (θ)}{\sin (θ)})$

Cancel the common factor of $\sin (θ)$ .

Cancel the common factor.

$1 = \frac{- 2}{1} (\frac{\sin (θ)}{1} \cdot \frac{\cos (θ)}{\sin (θ)})$

Rewrite the expression.

$1 = \frac{- 2}{1} \cos (θ)$

Divide $- 2$ by $1$ .

$1 = - 2 \cos (θ)$

Step 2

Rewrite the equation as $- 2 \cos (θ) = 1$ .

$- 2 \cos (θ) = 1$

Step 3

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

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

$\frac{- 2 \cos (θ)}{- 2} = \frac{1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 \cos (θ)}{- 2} = \frac{1}{- 2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$\cos (θ) = - \frac{1}{2}$

Step 4

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

$θ = \arccos (- \frac{1}{2})$

Step 5

Simplify the right side.

The exact value of $\arccos (- \frac{1}{2})$ is $\frac{2 π}{3}$ .

$θ = \frac{2 π}{3}$

Step 6

The cosine function is negative in the second and third quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the third quadrant.

$θ = 2 π - \frac{2 π}{3}$

Step 7

Simplify $2 π - \frac{2 π}{3}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$θ = 2 π \cdot \frac{3}{3} - \frac{2 π}{3}$

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

$θ = \frac{2 π \cdot 3}{3} - \frac{2 π}{3}$

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - 2 π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$θ = \frac{6 π - 2 π}{3}$

Subtract $2 π$ from $6 π$ .

$θ = \frac{4 π}{3}$

Step 8

Find the period of $\cos (θ)$ .

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Step 9

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

$θ = \frac{2 π}{3} + 2 π n, \frac{4 π}{3} + 2 π n$ , for any integer $n$

Solve for θ tan(theta)=-2sin(theta)

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