Trigonometry Examples

$2 \sin^{2} (θ) - 3 \sin (θ) + 1 = 0$ , $[0, 2 π)$

Step 1

Factor the left side of the equation.

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Step 1.1

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

$2 u^{2} - 3 u + 1 = 0$

Step 1.2

Factor by grouping.

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Step 1.2.1

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 $b = - 3$ .

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Step 1.2.1.1

Factor $- 3$ out of $- 3 u$ .

$2 u^{2} - 3 u + 1 = 0$

Step 1.2.1.2

Rewrite $- 3$ as $- 1$ plus $- 2$

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

Step 1.2.1.3

Apply the distributive property.

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

Step 1.2.2

Factor out the greatest common factor from each group.

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Step 1.2.2.1

Group the first two terms and the last two terms.

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

Step 1.2.2.2

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

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

Step 1.2.3

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

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

Step 1.3

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 $θ$ .

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Step 3.1

Set $2 \sin (θ) - 1$ equal to $0$ .

$2 \sin (θ) - 1 = 0$

Step 3.2

Solve $2 \sin (θ) - 1 = 0$ for $θ$ .

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Step 3.2.1

Add $1$ to both sides of the equation.

$2 \sin (θ) = 1$

Step 3.2.2

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

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Step 3.2.2.1

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

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

$\frac{2 \sin (θ)}{2} = \frac{1}{2}$

Step 3.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{1}{2}$

Step 3.2.3

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

$θ = \arcsin (\frac{1}{2})$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

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

$θ = \frac{π}{6}$

Step 3.2.5

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.

$θ = π - \frac{π}{6}$

Step 3.2.6

Simplify $π - \frac{π}{6}$ .

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Step 3.2.6.1

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

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

Step 3.2.6.2

Combine fractions.

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Step 3.2.6.2.1

Combine $π$ and $\frac{6}{6}$ .

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

Step 3.2.6.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 6 - π}{6}$

Step 3.2.6.3

Simplify the numerator.

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Step 3.2.6.3.1

Move $6$ to the left of $π$ .

$θ = \frac{6 \cdot π - π}{6}$

Step 3.2.6.3.2

Subtract $π$ from $6 π$ .

$θ = \frac{5 π}{6}$

Step 3.2.7

Find the period of $\sin (θ)$ .

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Step 3.2.7.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 3.2.7.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 3.2.7.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 3.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.8

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 $θ$ .

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Step 4.1

Set $\sin (θ) - 1$ equal to $0$ .

$\sin (θ) - 1 = 0$

Step 4.2

Solve $\sin (θ) - 1 = 0$ for $θ$ .

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Step 4.2.1

Add $1$ to both sides of the equation.

$\sin (θ) = 1$

Step 4.2.2

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

$θ = \arcsin (1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$θ = \frac{π}{2}$

Step 4.2.4

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.

$θ = π - \frac{π}{2}$

Step 4.2.5

Simplify $π - \frac{π}{2}$ .

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Step 4.2.5.1

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

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

Step 4.2.5.2

Combine fractions.

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Step 4.2.5.2.1

Combine $π$ and $\frac{2}{2}$ .

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

Step 4.2.5.2.2

Combine the numerators over the common denominator.

$θ = \frac{π \cdot 2 - π}{2}$

Step 4.2.5.3

Simplify the numerator.

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Step 4.2.5.3.1

Move $2$ to the left of $π$ .

$θ = \frac{2 \cdot π - π}{2}$

Step 4.2.5.3.2

Subtract $π$ from $2 π$ .

$θ = \frac{π}{2}$

Step 4.2.6

Find the period of $\sin (θ)$ .

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Step 4.2.6.1

The period of the function can be calculated using $\frac{2 π}{| b |}$ .

$\frac{2 π}{| b |}$

Step 4.2.6.2

Replace $b$ with $1$ in the formula for period.

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{2 π}{1}$

Step 4.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.7

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 5

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

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

Step 6

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

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Step 6.1

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

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Step 6.1.1

Plug in $0$ for $n$ .

$\frac{π}{6} + 2 π (0)$

Step 6.1.2

Simplify.

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Step 6.1.2.1

Multiply $2 π (0)$ .

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Step 6.1.2.1.1

Multiply $0$ by $2$ .

$\frac{π}{6} + 0 π$

Step 6.1.2.1.2

Multiply $0$ by $π$ .

$\frac{π}{6} + 0$

Step 6.1.2.2

Add $\frac{π}{6}$ and $0$ .

$\frac{π}{6}$

Step 6.1.3

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

$θ = \frac{π}{6}$

Step 6.2

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

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Step 6.2.1

Plug in $0$ for $n$ .

$\frac{π}{2} + 2 π (0)$

Step 6.2.2

Simplify.

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Step 6.2.2.1

Multiply $2 π (0)$ .

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Step 6.2.2.1.1

Multiply $0$ by $2$ .

$\frac{π}{2} + 0 π$

Step 6.2.2.1.2

Multiply $0$ by $π$ .

$\frac{π}{2} + 0$

Step 6.2.2.2

Add $\frac{π}{2}$ and $0$ .

$\frac{π}{2}$

Step 6.2.3

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

$θ = \frac{π}{6}, \frac{π}{2}$

Step 6.3

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

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Step 6.3.1

Plug in $0$ for $n$ .

$\frac{5 π}{6} + 2 π (0)$

Step 6.3.2

Simplify.

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Step 6.3.2.1

Multiply $2 π (0)$ .

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Step 6.3.2.1.1

Multiply $0$ by $2$ .

$\frac{5 π}{6} + 0 π$

Step 6.3.2.1.2

Multiply $0$ by $π$ .

$\frac{5 π}{6} + 0$

Step 6.3.2.2

Add $\frac{5 π}{6}$ and $0$ .

$\frac{5 π}{6}$

Step 6.3.3

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

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