Solve for θ sin(theta)^2=5(cos(theta)+1)

$\sin^{2} (θ) = 5 (\cos (θ) + 1)$

Step 1

Subtract $5 (\cos (θ) + 1)$ from both sides of the equation.

$\sin^{2} (θ) - 5 (\cos (θ) + 1) = 0$

Step 2

Simplify each term.

Apply the distributive property.

$\sin^{2} (θ) - 5 \cos (θ) - 5 \cdot 1 = 0$

Multiply $- 5$ by $1$ .

$\sin^{2} (θ) - 5 \cos (θ) - 5 = 0$

Step 3

Replace $\sin^{2} (θ)$ with $1 - \cos^{2} (θ)$ .

$(1 - \cos^{2} (θ)) - 5 \cos (θ) - 5 = 0$

Step 4

Solve for $θ$ .

Subtract $5$ from $1$ .

$- 4 - \cos^{2} (θ) - 5 \cos (θ) = 0$

Factor by grouping.

Reorder terms.

$- \cos^{2} (θ) - 5 \cos (θ) - 4 = 0$

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot - 4 = 4$ and whose sum is $b = - 5$ .

Factor $- 5$ out of $- 5 \cos (θ)$ .

$- \cos^{2} (θ) - 5 \cos (θ) - 4 = 0$

Rewrite $- 5$ as $- 1$ plus $- 4$

$- \cos^{2} (θ) + (- 1 - 4) \cos (θ) - 4 = 0$

Apply the distributive property.

$- \cos^{2} (θ) - 1 \cos (θ) - 4 \cos (θ) - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- \cos^{2} (θ) - 1 \cos (θ) - 4 \cos (θ) - 4 = 0$

Factor out the greatest common factor (GCF) from each group.

$\cos (θ) (- \cos (θ) - 1) + 4 (- \cos (θ) - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $- \cos (θ) - 1$ .

$(- \cos (θ) - 1) (\cos (θ) + 4) = 0$

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

$- \cos (θ) - 1 = 0$

$\cos (θ) + 4 = 0$

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

Set $- \cos (θ) - 1$ equal to $0$ .

$- \cos (θ) - 1 = 0$

Solve $- \cos (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$- \cos (θ) = 1$

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

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

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

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{\cos (θ)}{1} = \frac{1}{- 1}$

Divide $\cos (θ)$ by $1$ .

$\cos (θ) = \frac{1}{- 1}$

Simplify the right side.

Divide $1$ by $- 1$ .

$\cos (θ) = - 1$

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 $n$

Set $\cos (θ) + 4$ equal to $0$ and solve for $θ$ .

Set $\cos (θ) + 4$ equal to $0$ .

$\cos (θ) + 4 = 0$

Solve $\cos (θ) + 4 = 0$ for $θ$ .

Subtract $4$ from both sides of the equation.

$\cos (θ) = - 4$

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 4$ does not fall in this range, there is no solution.

No solution

The final solution is all the values that make $(- \cos (θ) - 1) (\cos (θ) + 4) = 0$ true.

$θ = π + 2 π n$ , for any integer $n$

Solve for θ sin(theta)^2=5(cos(theta)+1)

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