Precalculus Examples

$7 \sin^{2} (x) - 14 \sin (x) + 2 = - 5$

Step 1

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

$7 {(u)}^{2} - 14 u + 2 = - 5$

Step 2

Add $5$ to both sides of the equation.

$7 u^{2} - 14 u + 2 + 5 = 0$

Step 3

Add $2$ and $5$ .

$7 u^{2} - 14 u + 7 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $7$ out of $7 u^{2} - 14 u + 7$ .

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Step 4.1.1

Factor $7$ out of $7 u^{2}$ .

$7 u^{2} - 14 u + 7 = 0$

Step 4.1.2

Factor $7$ out of $- 14 u$ .

$7 u^{2} + 7 (- 2 u) + 7 = 0$

Step 4.1.3

Factor $7$ out of $7$ .

$7 u^{2} + 7 (- 2 u) + 7 (1) = 0$

Step 4.1.4

Factor $7$ out of $7 u^{2} + 7 (- 2 u)$ .

$7 (u^{2} - 2 u) + 7 (1) = 0$

Step 4.1.5

Factor $7$ out of $7 (u^{2} - 2 u) + 7 (1)$ .

$7 (u^{2} - 2 u + 1) = 0$

Step 4.2

Factor using the perfect square rule.

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Step 4.2.1

Rewrite $1$ as $1^{2}$ .

$7 (u^{2} - 2 u + 1^{2}) = 0$

Step 4.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 u = 2 \cdot u \cdot 1$

Step 4.2.3

Rewrite the polynomial.

$7 (u^{2} - 2 \cdot u \cdot 1 + 1^{2}) = 0$

Step 4.2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = u$ and $b = 1$ .

$7 {(u - 1)}^{2} = 0$

Step 5

Divide each term in $7 {(u - 1)}^{2} = 0$ by $7$ and simplify.

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Step 5.1

Divide each term in $7 {(u - 1)}^{2} = 0$ by $7$ .

$\frac{7 {(u - 1)}^{2}}{7} = \frac{0}{7}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $7$ .

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Step 5.2.1.1

Cancel the common factor.

$\frac{7 {(u - 1)}^{2}}{7} = \frac{0}{7}$

Step 5.2.1.2

Divide ${(u - 1)}^{2}$ by $1$ .

${(u - 1)}^{2} = \frac{0}{7}$

Step 5.3

Simplify the right side.

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Step 5.3.1

Divide $0$ by $7$ .

${(u - 1)}^{2} = 0$

Step 6

Set the $u - 1$ equal to $0$ .

$u - 1 = 0$

Step 7

Add $1$ to both sides of the equation.

$u = 1$

Step 8

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

$\sin (x) = 1$

Step 9

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

$x = \arcsin (1)$

Step 10

Simplify the right side.

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Step 10.1

The exact value of $\arcsin (1)$ is $\frac{π}{2}$ .

$x = \frac{π}{2}$

Step 11

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.

$x = π - \frac{π}{2}$

Step 12

Simplify $π - \frac{π}{2}$ .

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Step 12.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$x = π \cdot \frac{2}{2} - \frac{π}{2}$

Step 12.2

Combine fractions.

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Step 12.2.1

Combine $π$ and $\frac{2}{2}$ .

$x = \frac{π \cdot 2}{2} - \frac{π}{2}$

Step 12.2.2

Combine the numerators over the common denominator.

$x = \frac{π \cdot 2 - π}{2}$

Step 12.3

Simplify the numerator.

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Step 12.3.1

Move $2$ to the left of $π$ .

$x = \frac{2 \cdot π - π}{2}$

Step 12.3.2

Subtract $π$ from $2 π$ .

$x = \frac{π}{2}$

Step 13

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

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Step 13.1

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

$\frac{2 π}{| b |}$

Step 13.2

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

$\frac{2 π}{| 1 |}$

Step 13.3

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

$\frac{2 π}{1}$

Step 13.4

Divide $2 π$ by $1$ .

$2 π$

Step 14

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

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

Step 15