Trigonometry Examples

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

Step 1

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

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Step 1.1

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

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

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.

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

Step 1.2.1.2

Divide $\sin (x + \sqrt{2})$ by $1$ .

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

Step 1.3

Simplify the right side.

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Step 1.3.1

Divide $0$ by $2$ .

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

Step 2

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

$x + \sqrt{2} = \arcsin (0)$

Step 3

Simplify the right side.

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Step 3.1

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

$x + \sqrt{2} = 0$

Step 4

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

$x = - \sqrt{2}$

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.

$x + \sqrt{2} = π - 0$

Step 6

Solve for $x$ .

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Step 6.1

Subtract $0$ from $π$ .

$x + \sqrt{2} = π$

Step 6.2

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

$x = π - \sqrt{2}$

Step 7

Find the period of $\sin (x + \sqrt{2})$ .

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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 $- \sqrt{2}$ to find the positive angle.

$- \sqrt{2} + 2 π$

Step 8.2

List the new angles.

$x = - \sqrt{2} + 2 π$

Step 9

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

$x = - \sqrt{2} + 2 π + 2 π n, π - \sqrt{2} + 2 π + 2 π n$ , for any integer $n$