Trigonometry Examples

$2 \sin (θ) \cos (θ) + \sqrt{2} \cos (θ) = 0$

Step 1

Factor $\cos (θ)$ out of $2 \sin (θ) \cos (θ) + \sqrt{2} \cos (θ)$ .

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Step 1.1

Factor $\cos (θ)$ out of $2 \sin (θ) \cos (θ)$ .

$\cos (θ) (2 \sin (θ)) + \sqrt{2} \cos (θ) = 0$

Step 1.2

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

$\cos (θ) (2 \sin (θ)) + \cos (θ) \sqrt{2} = 0$

Step 1.3

Factor $\cos (θ)$ out of $\cos (θ) (2 \sin (θ)) + \cos (θ) \sqrt{2}$ .

$\cos (θ) (2 \sin (θ) + \sqrt{2}) = 0$

Step 2

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

$\cos (θ) = 0$

$2 \sin (θ) + \sqrt{2} = 0$

Step 3

Set $\cos (θ)$ equal to $0$ and solve for $θ$ .

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Step 3.1

Set $\cos (θ)$ equal to $0$ .

$\cos (θ) = 0$

Step 3.2

Solve $\cos (θ) = 0$ for $θ$ .

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Step 3.2.1

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

$θ = \arccos (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

The exact value of $\arccos (0)$ is $\frac{π}{2}$ .

$θ = \frac{π}{2}$

Step 3.2.3

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$θ = 2 π - \frac{π}{2}$

Step 3.2.4

Simplify $2 π - \frac{π}{2}$ .

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Step 3.2.4.1

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

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

Step 3.2.4.2

Combine fractions.

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Step 3.2.4.2.1

Combine $2 π$ and $\frac{2}{2}$ .

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

Step 3.2.4.2.2

Combine the numerators over the common denominator.

$θ = \frac{2 π \cdot 2 - π}{2}$

Step 3.2.4.3

Simplify the numerator.

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Step 3.2.4.3.1

Multiply $2$ by $2$ .

$θ = \frac{4 π - π}{2}$

Step 3.2.4.3.2

Subtract $π$ from $4 π$ .

$θ = \frac{3 π}{2}$

Step 3.2.5

Find the period of $\cos (θ)$ .

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Step 3.2.5.1

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

$\frac{2 π}{| b |}$

Step 3.2.5.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.5.3

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

$\frac{2 π}{1}$

Step 3.2.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.6

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

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

Step 4

Set $2 \sin (θ) + \sqrt{2}$ equal to $0$ and solve for $θ$ .

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Step 4.1

Set $2 \sin (θ) + \sqrt{2}$ equal to $0$ .

$2 \sin (θ) + \sqrt{2} = 0$

Step 4.2

Solve $2 \sin (θ) + \sqrt{2} = 0$ for $θ$ .

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Step 4.2.1

Subtract $\sqrt{2}$ from both sides of the equation.

$2 \sin (θ) = - \sqrt{2}$

Step 4.2.2

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

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Step 4.2.2.1

Divide each term in $2 \sin (θ) = - \sqrt{2}$ by $2$ .

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

Step 4.2.2.2

Simplify the left side.

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Step 4.2.2.2.1

Cancel the common factor of $2$ .

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Step 4.2.2.2.1.1

Cancel the common factor.

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

Step 4.2.2.2.1.2

Divide $\sin (θ)$ by $1$ .

$\sin (θ) = \frac{- \sqrt{2}}{2}$

Step 4.2.2.3

Simplify the right side.

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Step 4.2.2.3.1

Move the negative in front of the fraction.

$\sin (θ) = - \frac{\sqrt{2}}{2}$

Step 4.2.3

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

$θ = \arcsin (- \frac{\sqrt{2}}{2})$

Step 4.2.4

Simplify the right side.

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Step 4.2.4.1

The exact value of $\arcsin (- \frac{\sqrt{2}}{2})$ is $- \frac{π}{4}$ .

$θ = - \frac{π}{4}$

Step 4.2.5

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.

$θ = 2 π + \frac{π}{4} + π$

Step 4.2.6

Simplify the expression to find the second solution.

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Step 4.2.6.1

Subtract $2 π$ from $2 π + \frac{π}{4} + π$ .

$θ = 2 π + \frac{π}{4} + π - 2 π$

Step 4.2.6.2

The resulting angle of $\frac{5 π}{4}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{4} + π$ .

$θ = \frac{5 π}{4}$

Step 4.2.7

Find the period of $\sin (θ)$ .

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Step 4.2.7.1

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

$\frac{2 π}{| b |}$

Step 4.2.7.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.7.3

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

$\frac{2 π}{1}$

Step 4.2.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.8

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

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Step 4.2.8.1

Add $2 π$ to $- \frac{π}{4}$ to find the positive angle.

$- \frac{π}{4} + 2 π$

Step 4.2.8.2

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

$2 π \cdot \frac{4}{4} - \frac{π}{4}$

Step 4.2.8.3

Combine fractions.

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Step 4.2.8.3.1

Combine $2 π$ and $\frac{4}{4}$ .

$\frac{2 π \cdot 4}{4} - \frac{π}{4}$

Step 4.2.8.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 4 - π}{4}$

Step 4.2.8.4

Simplify the numerator.

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Step 4.2.8.4.1

Multiply $4$ by $2$ .

$\frac{8 π - π}{4}$

Step 4.2.8.4.2

Subtract $π$ from $8 π$ .

$\frac{7 π}{4}$

Step 4.2.8.5

List the new angles.

$θ = \frac{7 π}{4}$

Step 4.2.9

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

$θ = \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Step 5

The final solution is all the values that make $\cos (θ) (2 \sin (θ) + \sqrt{2}) = 0$ true.

$θ = \frac{π}{2} + 2 π n, \frac{3 π}{2} + 2 π n, \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Step 6

Consolidate $\frac{π}{2} + 2 π n$ and $\frac{3 π}{2} + 2 π n$ to $\frac{π}{2} + π n$ .

$θ = \frac{π}{2} + π n, \frac{5 π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$