Trigonometry Examples

$(\tan (x - \sqrt{3})) (2 \sin (x - 1)) = 0$

Step 1

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

$\tan (x - \sqrt{3}) = 0$

$\sin (x - 1) = 0$

Step 2

Set $\tan (x - \sqrt{3})$ equal to $0$ and solve for $x$ .

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Step 2.1

Set $\tan (x - \sqrt{3})$ equal to $0$ .

$\tan (x - \sqrt{3}) = 0$

Step 2.2

Solve $\tan (x - \sqrt{3}) = 0$ for $x$ .

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Step 2.2.1

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

$x - \sqrt{3} = \arctan (0)$

Step 2.2.2

Simplify the right side.

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Step 2.2.2.1

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

$x - \sqrt{3} = 0$

Step 2.2.3

Add $\sqrt{3}$ to both sides of the equation.

$x = \sqrt{3}$

Step 2.2.4

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$x - \sqrt{3} = π + 0$

Step 2.2.5

Solve for $x$ .

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Step 2.2.5.1

Add $π$ and $0$ .

$x - \sqrt{3} = π$

Step 2.2.5.2

Add $\sqrt{3}$ to both sides of the equation.

$x = π + \sqrt{3}$

Step 2.2.6

Find the period of $\tan (x - \sqrt{3})$ .

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Step 2.2.6.1

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

$\frac{π}{| b |}$

Step 2.2.6.2

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

$\frac{π}{| 1 |}$

Step 2.2.6.3

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

$\frac{π}{1}$

Step 2.2.6.4

Divide $π$ by $1$ .

$π$

Step 2.2.7

The period of the $\tan (x - \sqrt{3})$ function is $π$ so values will repeat every $π$ radians in both directions.

$x = \sqrt{3} + π n, π + \sqrt{3} + π n$ , for any integer $n$

Step 3

Set $\sin (x - 1)$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\sin (x - 1)$ equal to $0$ .

$\sin (x - 1) = 0$

Step 3.2

Solve $\sin (x - 1) = 0$ 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.

$x - 1 = \arcsin (0)$

Step 3.2.2

Simplify the right side.

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Step 3.2.2.1

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

$x - 1 = 0$

Step 3.2.3

Add $1$ to both sides of the equation.

$x = 1$

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.

$x - 1 = π - 0$

Step 3.2.5

Solve for $x$ .

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Step 3.2.5.1

Subtract $0$ from $π$ .

$x - 1 = π$

Step 3.2.5.2

Add $1$ to both sides of the equation.

$x = π + 1$

Step 3.2.6

Find the period of $\sin (x - 1)$ .

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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 $\sin (x - 1)$ 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 $n$

Step 4

The final solution is all the values that make $\tan (x - \sqrt{3}) \cdot (2 \sin (x - 1)) = 0$ true.

$x = \sqrt{3} + π n, π + \sqrt{3} + π n, 1 + 2 π n, π + 1 + 2 π n$ , for any integer $n$

Step 5

Consolidate the answers.

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Step 5.1

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

$x = \sqrt{3} + π n, 1 + 2 π n, π + 1 + 2 π n$ , for any integer $n$

Step 5.2

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

$x = \sqrt{3} + π n, 1 + π n$ , for any integer $n$

Step 5.3

Consolidate the answers.

$x = 1 + \sqrt{3} - 1 n$ , for any integer $n$

Step 6

Verify each of the solutions by substituting them into $(\tan (x - \sqrt{3})) (2 \sin (x - 1)) = 0$ and solving.

$x = \sqrt{3}$ , for any integer $n$