Solve for x sin(x-1)=0

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Trigonometry

$\sin (x - 1) = 0$

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

$x - 1 = \arcsin (0)$

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

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.

$x - 1 = π - 0$

Simplify the expression to find the second solution.

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Subtract $0$ from $π$ .

$x - 1 = π$

Add $1$ to both sides of the equation.

$x = π + 1$

Find the period.

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

Solve the equation.

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

The period of the $\sin (x - 1)$ function is $2 π$ so values will repeat every $2 π$ radians in both directions.

$x = 1 + 2 π n, π + 1 + 2 π n$ , for any integer $n$

Consolidate $1 + 2 π n$ and $π + 1 + 2 π n$ to $1 + π n$ .

$x = 1 + π n$ , for any integer $n$

Verify each of the solutions by substituting them into $\sin (x - 1) = 0$ and solving.

$x = 1$ , for any integer $n$

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