Solve for θ tan(theta)*sin(theta)=-sin(theta)

$\tan (θ) \cdot \sin (θ) = - \sin (θ)$

Step 1

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

$\tan (θ) \cdot \sin (θ) + \sin (θ) = 0$

Step 2

Simplify the left side of the equation.

Simplify each term.

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

$\frac{\sin (θ)}{\cos (θ)} \cdot \sin (θ) + \sin (θ) = 0$

Multiply $\frac{\sin (θ)}{\cos (θ)} \sin (θ)$ .

Combine $\frac{\sin (θ)}{\cos (θ)}$ and $\sin (θ)$ .

$\frac{\sin (θ) \sin (θ)}{\cos (θ)} + \sin (θ) = 0$

Raise $\sin (θ)$ to the power of $1$ .

$\frac{\sin (θ) \sin (θ)}{\cos (θ)} + \sin (θ) = 0$

Raise $\sin (θ)$ to the power of $1$ .

$\frac{\sin (θ) \sin (θ)}{\cos (θ)} + \sin (θ) = 0$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{{\sin (θ)}^{1 + 1}}{\cos (θ)} + \sin (θ) = 0$

Add $1$ and $1$ .

$\frac{\sin^{2} (θ)}{\cos (θ)} + \sin (θ) = 0$

Simplify each term.

Factor $\sin (θ)$ out of $\sin^{2} (θ)$ .

$\frac{\sin (θ) \sin (θ)}{\cos (θ)} + \sin (θ) = 0$

Separate fractions.

$\frac{\sin (θ)}{1} \cdot \frac{\sin (θ)}{\cos (θ)} + \sin (θ) = 0$

Convert from $\frac{\sin (θ)}{\cos (θ)}$ to $\tan (θ)$ .

$\frac{\sin (θ)}{1} \cdot \tan (θ) + \sin (θ) = 0$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) \tan (θ) + \sin (θ) = 0$

Step 3

Factor $\sin (θ)$ out of $\sin (θ) \tan (θ) + \sin (θ)$ .

Factor $\sin (θ)$ out of $\sin^{1} (θ)$ .

$\sin (θ) \tan (θ) + \sin (θ) \cdot 1 = 0$

Factor $\sin (θ)$ out of $\sin (θ) \tan (θ) + \sin (θ) \cdot 1$ .

$\sin (θ) (\tan (θ) + 1) = 0$

Step 4

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

$\sin (θ) = 0$

$\tan (θ) + 1 = 0$

Step 5

Set $\sin (θ)$ equal to $0$ and solve for $θ$ .

Set $\sin (θ)$ equal to $0$ .

$\sin (θ) = 0$

Solve $\sin (θ) = 0$ for $θ$ .

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

$θ = \arcsin (0)$

Simplify the right side.

The exact value of $\arcsin (0)$ is $0$ .

$θ = 0$

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

$θ = π - 0$

Subtract $0$ from $π$ .

$θ = π$

Find the period of $\sin (θ)$ .

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 $1$ .

$2 π$

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

$θ = 2 π n, π + 2 π n$ , for any integer $n$

Step 6

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

Set $\tan (θ) + 1$ equal to $0$ .

$\tan (θ) + 1 = 0$

Solve $\tan (θ) + 1 = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$\tan (θ) = - 1$

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

$θ = \arctan (- 1)$

Simplify the right side.

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

$θ = - \frac{π}{4}$

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.

$θ = - \frac{π}{4} - π$

Simplify the expression to find the second solution.

Add $2 π$ to $- \frac{π}{4} - π$ .

$θ = - \frac{π}{4} - π + 2 π$

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

$θ = \frac{3 π}{4}$

Find the period of $\tan (θ)$ .

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

$\frac{π}{| b |}$

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

$\frac{π}{| 1 |}$

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

$\frac{π}{1}$

Divide $π$ by $1$ .

$π$

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

Add $π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + π$

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

$π \cdot \frac{4}{4} - \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$\frac{π \cdot 4}{4} - \frac{π}{4}$

Combine the numerators over the common denominator.

$\frac{π \cdot 4 - π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$\frac{4 \cdot π - π}{4}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{4}$

List the new angles.

$θ = \frac{3 π}{4}$

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

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

Step 7

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

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

Step 8

Consolidate $2 π n$ and $π + 2 π n$ to $π n$ .

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

Solve for θ tan(theta)*sin(theta)=-sin(theta)

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