Trigonometry Examples

$5 x \sin (x - 5) = 0$

Step 1

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

$x = 0$

$\sin (x - 5) = 0$

Step 2

Set $x$ equal to $0$ .

$x = 0$

Step 3

Set $\sin (x - 5)$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\sin (x - 5)$ equal to $0$ .

$\sin (x - 5) = 0$

Step 3.2

Solve $\sin (x - 5) = 0$ for $x$ .

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Step 3.2.1

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

$x - 5 = \arcsin (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$x - 5 = 0$

Step 3.2.3

Add $5$ to both sides of the equation.

$x = 5$

Step 3.2.4

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.

$x - 5 = π - 0$

Step 3.2.5

Solve for $x$ .

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Step 3.2.5.1

Subtract $0$ from $π$ .

$x - 5 = π$

Step 3.2.5.2

Add $5$ to both sides of the equation.

$x = π + 5$

Step 3.2.6

Find the period of $\sin (x - 5)$ .

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Step 3.2.6.1

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

$\frac{2 π}{| b |}$

Step 3.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 3.2.6.3

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

$\frac{2 π}{1}$

Step 3.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.2.7

The period of the $\sin (x - 5)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 5 + 2 π n, π + 5 + 2 π n$ , for any integer $n$

Step 4

The final solution is all the values that make $5 x \sin (x - 5) = 0$ true.

$x = 0, 5 + 2 π n, π + 5 + 2 π n$ , for any integer $n$