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

$2 \cos^{2} (θ) = 2 + \sin (θ)$

Step 1

Move all the expressions to the left side of the equation.

Subtract $2$ from both sides of the equation.

$2 \cos^{2} (θ) - 2 = \sin (θ)$

Subtract $\sin (θ)$ from both sides of the equation.

$2 \cos^{2} (θ) - 2 - \sin (θ) = 0$

Step 2

Simplify $2 \cos^{2} (θ) - 2 - \sin (θ)$ .

Reorder $2 \cos^{2} (θ)$ and $- 2$ .

$- 2 + 2 \cos^{2} (θ) - \sin (θ) = 0$

Factor $- 2$ out of $- 2$ .

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

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

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

Factor $- 2$ out of $- 2 (1) - 2 (- \cos^{2} (θ))$ .

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

Apply pythagorean identity.

$- 2 \sin^{2} (θ) - \sin (θ) = 0$

Step 3

Solve for $θ$ .

Factor the left side of the equation.

Let $u = \sin (θ)$ . Substitute $u$ for all occurrences of $\sin (θ)$ .

$- 2 u^{2} - u = 0$

Factor $u$ out of $- 2 u^{2} - u$ .

Factor $u$ out of $- 2 u^{2}$ .

$u (- 2 u) - u = 0$

Factor $u$ out of $- u$ .

$u (- 2 u) + u \cdot - 1 = 0$

Factor $u$ out of $u (- 2 u) + u \cdot - 1$ .

$u (- 2 u - 1) = 0$

Replace all occurrences of $u$ with $\sin (θ)$ .

$\sin (θ) (- 2 \sin (θ) - 1) = 0$

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

$\sin (θ) = 0$

$- 2 \sin (θ) - 1 = 0$

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 $- 2 \sin (θ) - 1$ equal to $0$ and solve for $θ$ .

Set $- 2 \sin (θ) - 1$ equal to $0$ .

$- 2 \sin (θ) - 1 = 0$

Solve $- 2 \sin (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$- 2 \sin (θ) = 1$

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

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

$\frac{- 2 \sin (θ)}{- 2} = \frac{1}{- 2}$

Simplify the left side.

Cancel the common factor of $- 2$ .

Cancel the common factor.

$\frac{- 2 \sin (θ)}{- 2} = \frac{1}{- 2}$

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{- 2}$

Simplify the right side.

Move the negative in front of the fraction.

$\sin (θ) = - \frac{1}{2}$

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

$θ = \arcsin (- \frac{1}{2})$

Simplify the right side.

The exact value of $\arcsin (- \frac{1}{2})$ is $- \frac{π}{6}$ .

$θ = - \frac{π}{6}$

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$θ = 2 π + \frac{π}{6} + π$

Simplify the expression to find the second solution.

Subtract $2 π$ from $2 π + \frac{π}{6} + π$ .

$θ = 2 π + \frac{π}{6} + π - 2 π$

The resulting angle of $\frac{7 π}{6}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{6} + π$ .

$θ = \frac{7 π}{6}$

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 $1$ .

$2 π$

Add $2 π$ to every negative angle to get positive angles.

Add $2 π$ to $- \frac{π}{6}$ to find the positive angle.

$- \frac{π}{6} + 2 π$

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{6}{6}$ .

$2 π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $2 π$ and $\frac{6}{6}$ .

$\frac{2 π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 6 - π}{6}$

Simplify the numerator.

Multiply $6$ by $2$ .

$\frac{12 π - π}{6}$

Subtract $π$ from $12 π$ .

$\frac{11 π}{6}$

List the new angles.

$θ = \frac{11 π}{6}$

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

$θ = \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

The final solution is all the values that make $\sin (θ) (- 2 \sin (θ) - 1) = 0$ true.

$θ = 2 π n, π + 2 π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

Step 4

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

$θ = π n, \frac{7 π}{6} + 2 π n, \frac{11 π}{6} + 2 π n$ , for any integer $n$

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

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