Solve Over the Interval 2sin(theta)^2+sin(theta)-1=0 , [0,2pi)

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

Step 1

Factor the left side of the equation.

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

$2 u^{2} + u - 1 = 0$

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 $b = 1$ .

Multiply by $1$ .

$2 u^{2} + 1 u - 1 = 0$

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

$2 u^{2} + (- 1 + 2) u - 1 = 0$

Apply the distributive property.

$2 u^{2} - 1 u + 2 u - 1 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(2 u^{2} - 1 u) + 2 u - 1 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

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

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

$(2 \sin (θ) - 1) (\sin (θ) + 1) = 0$

Step 2

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 3

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 $1$ .

$\sin (θ) = \frac{1}{2}$

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 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.

$θ = π - \frac{π}{6}$

Simplify $π - \frac{π}{6}$ .

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

$θ = π \cdot \frac{6}{6} - \frac{π}{6}$

Combine fractions.

Combine $π$ and $\frac{6}{6}$ .

$θ = \frac{π \cdot 6}{6} - \frac{π}{6}$

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Simplify the numerator.

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

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 $1$ .

$2 π$

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

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n$ , for any integer $n$

Step 4

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 $θ$ .

Subtract $1$ from 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 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.

$θ = 2 π + \frac{π}{2} + π$

Simplify the expression to find the second solution.

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

$θ = 2 π + \frac{π}{2} + π - 2 π$

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$θ = \frac{3 π}{2}$

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 $1$ .

$2 π$

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

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

$- \frac{π}{2} + 2 π$

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Combine fractions.

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

List the new angles.

$θ = \frac{3 π}{2}$

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

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

Step 5

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

$θ = \frac{π}{6} + 2 π n, \frac{5 π}{6} + 2 π n, \frac{3 π}{2} + 2 π n$ , for any integer $n$

Step 6

Consolidate the answers.

$θ = \frac{π}{6} + \frac{2 π n}{3}$ , for any integer $n$

Step 7

Find the values of $n$ that produce a value within the interval $[0, 2 π)$ .

Plug in $0$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $0$ for $n$ .

$\frac{π}{6} + \frac{2 π (0)}{3}$

Simplify.

Simplify each term.

Cancel the common factor of $0$ and $3$ .

Factor $3$ out of $2 π (0)$ .

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3}$

Cancel the common factors.

Factor $3$ out of $3$ .

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3 (1)}$

Cancel the common factor.

$\frac{π}{6} + \frac{3 (2 π \cdot (0))}{3 \cdot 1}$

Rewrite the expression.

$\frac{π}{6} + \frac{2 π \cdot (0)}{1}$

Divide $2 π \cdot (0)$ by $1$ .

$\frac{π}{6} + 2 π \cdot (0)$

Multiply $2 π (0)$ .

Multiply $0$ by $2$ .

$\frac{π}{6} + 0 π$

Multiply $0$ by $π$ .

$\frac{π}{6} + 0$

Add $\frac{π}{6}$ and $0$ .

$\frac{π}{6}$

The interval $[0, 2 π)$ contains $\frac{π}{6}$ .

$θ = \frac{π}{6}$

Plug in $1$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $1$ for $n$ .

$\frac{π}{6} + \frac{2 π (1)}{3}$

Simplify.

Multiply $2$ by $1$ .

$\frac{π}{6} + \frac{2 π}{3}$

To write $\frac{2 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{2 π}{3} \cdot \frac{2}{2}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{2 π}{3}$ by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{2 π \cdot 2}{3 \cdot 2}$

Multiply $3$ by $2$ .

$\frac{π}{6} + \frac{2 π \cdot 2}{6}$

Combine the numerators over the common denominator.

$\frac{π + 2 π \cdot 2}{6}$

Simplify the numerator.

Multiply $2$ by $2$ .

$\frac{π + 4 π}{6}$

Add $π$ and $4 π$ .

$\frac{5 π}{6}$

The interval $[0, 2 π)$ contains $\frac{5 π}{6}$ .

$θ = \frac{π}{6}, \frac{5 π}{6}$

Plug in $2$ for $n$ and simplify to see if the solution is contained in $[0, 2 π)$ .

Plug in $2$ for $n$ .

$\frac{π}{6} + \frac{2 π (2)}{3}$

Simplify.

Multiply $2$ by $2$ .

$\frac{π}{6} + \frac{4 π}{3}$

To write $\frac{4 π}{3}$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{4 π}{3} \cdot \frac{2}{2}$

Write each expression with a common denominator of $6$ , by multiplying each by an appropriate factor of $1$ .

Multiply $\frac{4 π}{3}$ by $\frac{2}{2}$ .

$\frac{π}{6} + \frac{4 π \cdot 2}{3 \cdot 2}$

Multiply $3$ by $2$ .

$\frac{π}{6} + \frac{4 π \cdot 2}{6}$

Combine the numerators over the common denominator.

$\frac{π + 4 π \cdot 2}{6}$

Simplify the numerator.

Multiply $2$ by $4$ .

$\frac{π + 8 π}{6}$

Add $π$ and $8 π$ .

$\frac{9 π}{6}$

Cancel the common factor of $9$ and $6$ .

Factor $3$ out of $9 π$ .

$\frac{3 (3 π)}{6}$

Cancel the common factors.

Factor $3$ out of $6$ .

$\frac{3 (3 π)}{3 (2)}$

Cancel the common factor.

$\frac{3 (3 π)}{3 \cdot 2}$

Rewrite the expression.

$\frac{3 π}{2}$

The interval $[0, 2 π)$ contains $\frac{3 π}{2}$ .

$θ = \frac{π}{6}, \frac{5 π}{6}, \frac{3 π}{2}$

Solve Over the Interval 2sin(theta)^2+sin(theta)-1=0 , [0,2pi)

Related Questions

Support LCMGCF.com by disabling your adblocker.