Finite Math Examples

$6 \sin^{2} (x) - 4 \sin (x) = 1$

Step 1

Substitute $u$ for $\sin (x)$ .

$6 {(u)}^{2} - 4 u = 1$

Step 2

Subtract $1$ from both sides of the equation.

$6 u^{2} - 4 u - 1 = 0$

Step 3

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 4

Substitute the values $a = 6$ , $b = - 4$ , and $c = - 1$ into the quadratic formula and solve for $u$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (6 \cdot - 1)}}{2 \cdot 6}$

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise $- 4$ to the power of $2$ .

$u = \frac{4 \pm \sqrt{16 - 4 \cdot 6 \cdot - 1}}{2 \cdot 6}$

Step 5.1.2

Multiply $- 4 \cdot 6 \cdot - 1$ .

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Step 5.1.2.1

Multiply $- 4$ by $6$ .

$u = \frac{4 \pm \sqrt{16 - 24 \cdot - 1}}{2 \cdot 6}$

Step 5.1.2.2

Multiply $- 24$ by $- 1$ .

$u = \frac{4 \pm \sqrt{16 + 24}}{2 \cdot 6}$

Step 5.1.3

Add $16$ and $24$ .

$u = \frac{4 \pm \sqrt{40}}{2 \cdot 6}$

Step 5.1.4

Rewrite $40$ as $2^{2} \cdot 10$ .

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Step 5.1.4.1

Factor $4$ out of $40$ .

$u = \frac{4 \pm \sqrt{4 (10)}}{2 \cdot 6}$

Step 5.1.4.2

Rewrite $4$ as $2^{2}$ .

$u = \frac{4 \pm \sqrt{2^{2} \cdot 10}}{2 \cdot 6}$

Step 5.1.5

Pull terms out from under the radical.

$u = \frac{4 \pm 2 \sqrt{10}}{2 \cdot 6}$

Step 5.2

Multiply $2$ by $6$ .

$u = \frac{4 \pm 2 \sqrt{10}}{12}$

Step 5.3

Simplify $\frac{4 \pm 2 \sqrt{10}}{12}$ .

$u = \frac{2 \pm \sqrt{10}}{6}$

Step 6

The final answer is the combination of both solutions.

$u = \frac{2 + \sqrt{10}}{6}, \frac{2 - \sqrt{10}}{6}$

Step 7

Substitute $\sin (x)$ for $u$ .

$\sin (x) = \frac{2 + \sqrt{10}}{6}, \frac{2 - \sqrt{10}}{6}$

Step 8

Set up each of the solutions to solve for $x$ .

$\sin (x) = \frac{2 + \sqrt{10}}{6}$

$\sin (x) = \frac{2 - \sqrt{10}}{6}$

Step 9

Solve for $x$ in $\sin (x) = \frac{2 + \sqrt{10}}{6}$ .

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Step 9.1

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

$x = \arcsin (\frac{2 + \sqrt{10}}{6})$

Step 9.2

Simplify the right side.

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Step 9.2.1

Evaluate $\arcsin (\frac{2 + \sqrt{10}}{6})$ .

$x = 1.03601404$

Step 9.3

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.

$x = (3.14159265) - 1.03601404$

Step 9.4

Solve for $x$ .

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Step 9.4.1

Remove parentheses.

$x = 3.14159265 - 1.03601404$

Step 9.4.2

Remove parentheses.

$x = (3.14159265) - 1.03601404$

Step 9.4.3

Subtract $1.03601404$ from $3.14159265$ .

$x = 2.1055786$

Step 9.5

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

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Step 9.5.1

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

$\frac{2 π}{| b |}$

Step 9.5.2

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

$\frac{2 π}{| 1 |}$

Step 9.5.3

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

$\frac{2 π}{1}$

Step 9.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 9.6

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

$x = 1.03601404 + 2 π n, 2.1055786 + 2 π n$ , for any integer $n$

Step 10

Solve for $x$ in $\sin (x) = \frac{2 - \sqrt{10}}{6}$ .

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Step 10.1

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

$x = \arcsin (\frac{2 - \sqrt{10}}{6})$

Step 10.2

Simplify the right side.

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Step 10.2.1

Evaluate $\arcsin (\frac{2 - \sqrt{10}}{6})$ .

$x = - 0.19494537$

Step 10.3

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.

$x = (3.14159265) + 0.19494537$

Step 10.4

Solve for $x$ .

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Step 10.4.1

Remove parentheses.

$x = 3.14159265 + 0.19494537$

Step 10.4.2

Remove parentheses.

$x = (3.14159265) + 0.19494537$

Step 10.4.3

Add $3.14159265$ and $0.19494537$ .

$x = 3.33653802$

Step 10.5

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

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Step 10.5.1

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

$\frac{2 π}{| b |}$

Step 10.5.2

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

$\frac{2 π}{| 1 |}$

Step 10.5.3

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

$\frac{2 π}{1}$

Step 10.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 10.6

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

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Step 10.6.1

Add $2 π$ to $- 0.19494537$ to find the positive angle.

$- 0.19494537 + 2 π$

Step 10.6.2

Subtract $0.19494537$ from $2 π$ .

$6.08823993$

Step 10.6.3

List the new angles.

$x = 6.08823993$

Step 10.7

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

$x = 3.33653802 + 2 π n, 6.08823993 + 2 π n$ , for any integer $n$

Step 11

List all of the solutions.

$x = 1.03601404 + 2 π n, 2.1055786 + 2 π n, 3.33653802 + 2 π n, 6.08823993 + 2 π n$ , for any integer $n$