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

$2 \sin^{2} (θ) = \sin (θ) + 1$

Step 1

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

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

$2 \sin^{2} (θ) - \sin (θ) = 1$

Subtract $1$ from both sides of the equation.

$2 \sin^{2} (θ) - \sin (θ) - 1 = 0$

Step 2

Factor by grouping.

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 $a \cdot c = 2 \cdot - 1 = - 2$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- \sin (θ)$ .

$2 \sin^{2} (θ) - \sin (θ) - 1 = 0$

Rewrite $- 1$ as $1$ plus $- 2$

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

Apply the distributive property.

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

Multiply $\sin (θ)$ by $1$ .

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

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

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

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

$\sin (θ) (2 \sin (θ) + 1) - (2 \sin (θ) + 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $2 \sin (θ) + 1$ .

$(2 \sin (θ) + 1) (\sin (θ) - 1) = 0$

Step 3

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

$2 \sin (θ) + 1 = 0$

$\sin (θ) - 1 = 0$

Step 4

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 $θ$ .

Subtract $1$ from 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 $n$

Step 5

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

Set $\sin (θ) - 1$ equal to $0$ .

$\sin (θ) - 1 = 0$

Solve $\sin (θ) - 1 = 0$ for $θ$ .

Add $1$ to both sides of the equation.

$\sin (θ) = 1$

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

$θ = \arcsin (1)$

Simplify the right side.

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

$θ = \frac{π}{2}$

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.

$θ = π - \frac{π}{2}$

Simplify $π - \frac{π}{2}$ .

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

$θ = π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $π$ and $\frac{2}{2}$ .

$θ = \frac{π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Simplify the numerator.

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

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.

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

Step 6

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

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

Step 7

Consolidate the answers.

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

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

Related Questions

Support LCMGCF.com by disabling your adblocker.