Solve for θ in Radians tan(theta)=2sin(theta)

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Trigonometry

$\tan (θ) = 2 \sin (θ)$

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

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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.

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

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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.

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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 $\sin (θ)$ .

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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 $1$ .

$1 = 2 \cos (θ)$

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

$2 \cos (θ) = 1$

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

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

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

Simplify the left side.

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Cancel the common factor of $2$ .

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Cancel the common factor.

$\frac{2 \cos (θ)}{2} = \frac{1}{2}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{2}$

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

$θ = \arccos (\frac{1}{2})$

Simplify the right side.

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

$θ = \frac{π}{3}$

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

$θ = 2 π - \frac{π}{3}$

Simplify $2 π - \frac{π}{3}$ .

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

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

Combine fractions.

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Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

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Multiply $3$ by $2$ .

$θ = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{3}$

Find the period of $\cos (θ)$ .

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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 $1$ .

$2 π$

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

$θ = \frac{π}{3} + 2 π n, \frac{5 π}{3} + 2 π n$ , for any integer $n$

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