Solve for θ (1+cos(theta))(tan(theta)-1)=0

$(1 + \cos (θ)) (\tan (θ) - 1) = 0$

Step 1

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

$1 + \cos (θ) = 0$

$\tan (θ) - 1 = 0$

Step 2

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

Set $1 + \cos (θ)$ equal to $0$ .

$1 + \cos (θ) = 0$

Solve $1 + \cos (θ) = 0$ for $θ$ .

Subtract $1$ from both sides of the equation.

$\cos (θ) = - 1$

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

$θ = \arccos (- 1)$

Simplify the right side.

The exact value of $\arccos (- 1)$ is $π$ .

$θ = π$

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.

$θ = 2 π - π$

Subtract $π$ from $2 π$ .

$θ = π$

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 $1$ .

$2 π$

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

$θ = π + 2 π n$ , for any integer $n$

Step 3

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 $θ$ .

Add $1$ to 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 positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$θ = π + \frac{π}{4}$

Simplify $π + \frac{π}{4}$ .

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

$θ = π \cdot \frac{4}{4} + \frac{π}{4}$

Combine fractions.

Combine $π$ and $\frac{4}{4}$ .

$θ = \frac{π \cdot 4}{4} + \frac{π}{4}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 4 + π}{4}$

Simplify the numerator.

Move $4$ to the left of $π$ .

$θ = \frac{4 \cdot π + π}{4}$

Add $4 π$ and $π$ .

$θ = \frac{5 π}{4}$

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 $1$ .

$π$

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

$θ = \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 4

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

$θ = π + 2 π n, \frac{π}{4} + π n, \frac{5 π}{4} + π n$ , for any integer $n$

Step 5

Consolidate $\frac{π}{4} + π n$ and $\frac{5 π}{4} + π n$ to $\frac{π}{4} + π n$ .

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

Solve for θ (1+cos(theta))(tan(theta)-1)=0

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