Trigonometry Examples

$2 \sin^{2} (θ) - \sin (θ) - 1 = 0$

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 $\sin (θ)$ .

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

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 $b = - 1$ .

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Step 1.2.1.1

Factor $- 1$ out of $- u$ .

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

Step 1.2.1.2

Rewrite $- 1$ 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 1.2.1.4

Multiply $u$ by $1$ .

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

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.

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

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

Subtract $1$ from 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 3.2.2.3

Simplify the right side.

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Step 3.2.2.3.1

Move the negative in front of the fraction.

$\sin (θ) = - \frac{1}{2}$

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 $- 30$ .

$θ = - 30$

Step 3.2.5

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $360$ , to find a reference angle. Next, add this reference angle to $180$ to find the solution in the third quadrant.

$θ = 360 + 30 + 180$

Step 3.2.6

Simplify the expression to find the second solution.

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Step 3.2.6.1

Subtract $360 °$ from $360 + 30 + 180 °$ .

$θ = 360 + 30 + 180 ° - 360 °$

Step 3.2.6.2

The resulting angle of $210 °$ is positive, less than $360 °$ , and coterminal with $360 + 30 + 180$ .

$θ = 210 °$

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 $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 3.2.7.2

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

$\frac{360}{| 1 |}$

Step 3.2.7.3

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

$\frac{360}{1}$

Step 3.2.7.4

Divide $360$ by $1$ .

$360$

Step 3.2.8

Add $360$ to every negative angle to get positive angles.

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Step 3.2.8.1

Add $360$ to $- 30$ to find the positive angle.

$- 30 + 360$

Step 3.2.8.2

Subtract $30$ from $360$ .

$330$

Step 3.2.8.3

List the new angles.

$θ = 330$

Step 3.2.9

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 210 + 360 n, 330 + 360 n$ , 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 $90$ .

$θ = 90$

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 $180$ to find the solution in the second quadrant.

$θ = 180 - 90$

Step 4.2.5

Subtract $90$ from $180$ .

$θ = 90$

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 $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Step 4.2.6.2

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

$\frac{360}{| 1 |}$

Step 4.2.6.3

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

$\frac{360}{1}$

Step 4.2.6.4

Divide $360$ by $1$ .

$360$

Step 4.2.7

The period of the $\sin (θ)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$θ = 90 + 360 n$ , for any integer $n$

Step 5

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

$θ = 210 + 360 n, 330 + 360 n, 90 + 360 n$ , for any integer $n$

Step 6

Consolidate the answers.

$θ = 90 + 120 n$ , for any integer $n$