Trigonometry Examples

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

Step 1

Divide each term in $2 \sqrt{2} \sin (q + 2) = 0$ by $2 \sqrt{2}$ and simplify.

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Step 1.1

Divide each term in $2 \sqrt{2} \sin (q + 2) = 0$ by $2 \sqrt{2}$ .

$\frac{2 \sqrt{2} \sin (q + 2)}{2 \sqrt{2}} = \frac{0}{2 \sqrt{2}}$

Step 1.2

Simplify the left side.

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Step 1.2.1

Cancel the common factor of $2$ .

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Step 1.2.1.1

Cancel the common factor.

$\frac{2 \sqrt{2} \sin (q + 2)}{2 \sqrt{2}} = \frac{0}{2 \sqrt{2}}$

Step 1.2.1.2

Rewrite the expression.

$\frac{\sqrt{2} \sin (q + 2)}{\sqrt{2}} = \frac{0}{2 \sqrt{2}}$

Step 1.2.2

Cancel the common factor of $\sqrt{2}$ .

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Step 1.2.2.1

Cancel the common factor.

$\frac{\sqrt{2} \sin (q + 2)}{\sqrt{2}} = \frac{0}{2 \sqrt{2}}$

Step 1.2.2.2

Divide $\sin (q + 2)$ by $1$ .

$\sin (q + 2) = \frac{0}{2 \sqrt{2}}$

Step 1.3

Simplify the right side.

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Step 1.3.1

Cancel the common factor of $0$ and $2$ .

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Step 1.3.1.1

Factor $2$ out of $0$ .

$\sin (q + 2) = \frac{2 (0)}{2 \sqrt{2}}$

Step 1.3.1.2

Cancel the common factors.

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Step 1.3.1.2.1

Factor $2$ out of $2 \sqrt{2}$ .

$\sin (q + 2) = \frac{2 (0)}{2 (\sqrt{2})}$

Step 1.3.1.2.2

Cancel the common factor.

$\sin (q + 2) = \frac{2 \cdot 0}{2 \sqrt{2}}$

Step 1.3.1.2.3

Rewrite the expression.

$\sin (q + 2) = \frac{0}{\sqrt{2}}$

Step 1.3.2

Divide $0$ by $\sqrt{2}$ .

$\sin (q + 2) = 0$

Step 2

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

$q + 2 = \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

The exact value of $\arcsin (0)$ is $0$ .

$q + 2 = 0$

Step 4

Subtract $2$ from both sides of the equation.

$q = - 2$

Step 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.

$q + 2 = π - 0$

Step 6

Solve for $q$ .

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Step 6.1

Subtract $0$ from $π$ .

$q + 2 = π$

Step 6.2

Subtract $2$ from both sides of the equation.

$q = π - 2$

Step 7

Find the period of $\sin (q + 2)$ .

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Step 7.1

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

$\frac{2 π}{| b |}$

Step 7.2

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

$\frac{2 π}{| 1 |}$

Step 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 7.4

Divide $2 π$ by $1$ .

$2 π$

Step 8

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

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Step 8.1

Add $2 π$ to $- 2$ to find the positive angle.

$- 2 + 2 π$

Step 8.2

List the new angles.

$q = - 2 + 2 π$

Step 9

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

$q = - 2 + 2 π + 2 π n, π - 2 + 2 π + 2 π n$ , for any integer $n$