Solve for θ sin(theta)+cos(theta)tan(theta)^2=0

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

Simplify the left side of the equation.

Simplify each term.

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

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

Apply the product rule to $\frac{\sin (θ)}{\cos (θ)}$ .

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

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

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

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

Cancel the common factor.

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

Rewrite the expression.

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

Simplify each term.

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

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

Separate fractions.

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

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

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

Divide $\sin (θ)$ by $1$ .

$\sin (θ) + \sin (θ) \tan (θ) = 0$

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

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

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

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

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

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

$\sin (θ) (1 + \tan (θ)) = 0$

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

$\sin (θ) = 0$

$1 + \tan (θ) = 0$

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 $n$

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

Set $1 + \tan (θ)$ equal to $0$ .

$1 + \tan (θ) = 0$

Solve $1 + \tan (θ) = 0$ 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 $n$

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

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

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 sin(theta)+cos(theta)tan(theta)^2=0

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