Precalculus Examples

$\sin^{2} (x) - \sin (x) - 2 = 0$

Step 1

Factor the left side of the equation.

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Step 1.1

Let $u = \sin (x)$ . Substitute $u$ for all occurrences of $\sin (x)$ .

$u^{2} - u - 2 = 0$

Step 1.2

Factor $u^{2} - u - 2$ using the AC method.

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Step 1.2.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 2$ and whose sum is $- 1$ .

$- 2, 1$

Step 1.2.2

Write the factored form using these integers.

$(u - 2) (u + 1) = 0$

Step 1.3

Replace all occurrences of $u$ with $\sin (x)$ .

$(\sin (x) - 2) (\sin (x) + 1) = 0$

Step 2

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

$\sin (x) - 2 = 0$

$\sin (x) + 1 = 0$

Step 3

Set $\sin (x) - 2$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $\sin (x) - 2$ equal to $0$ .

$\sin (x) - 2 = 0$

Step 3.2

Solve $\sin (x) - 2 = 0$ for $x$ .

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Step 3.2.1

Add $2$ to both sides of the equation.

$\sin (x) = 2$

Step 3.2.2

The range of sine is $- 1 \leq y \leq 1$ . Since $2$ does not fall in this range, there is no solution.

No solution

Step 4

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

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Step 4.1

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

$\sin (x) + 1 = 0$

Step 4.2

Solve $\sin (x) + 1 = 0$ for $x$ .

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$\sin (x) = - 1$

Step 4.2.2

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

$x = \arcsin (- 1)$

Step 4.2.3

Simplify the right side.

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Step 4.2.3.1

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

$x = - \frac{π}{2}$

Step 4.2.4

The sine function is negative in the third and fourth quadrants. To find the second solution, subtract the solution from $2 π$ , to find a reference angle. Next, add this reference angle to $π$ to find the solution in the third quadrant.

$x = 2 π + \frac{π}{2} + π$

Step 4.2.5

Simplify the expression to find the second solution.

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Step 4.2.5.1

Subtract $2 π$ from $2 π + \frac{π}{2} + π$ .

$x = 2 π + \frac{π}{2} + π - 2 π$

Step 4.2.5.2

The resulting angle of $\frac{3 π}{2}$ is positive, less than $2 π$ , and coterminal with $2 π + \frac{π}{2} + π$ .

$x = \frac{3 π}{2}$

Step 4.2.6

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

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Step 4.2.6.1

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

$\frac{2 π}{| b |}$

Step 4.2.6.2

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

$\frac{2 π}{| 1 |}$

Step 4.2.6.3

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

$\frac{2 π}{1}$

Step 4.2.6.4

Divide $2 π$ by $1$ .

$2 π$

Step 4.2.7

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

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Step 4.2.7.1

Add $2 π$ to $- \frac{π}{2}$ to find the positive angle.

$- \frac{π}{2} + 2 π$

Step 4.2.7.2

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

$2 π \cdot \frac{2}{2} - \frac{π}{2}$

Step 4.2.7.3

Combine fractions.

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Step 4.2.7.3.1

Combine $2 π$ and $\frac{2}{2}$ .

$\frac{2 π \cdot 2}{2} - \frac{π}{2}$

Step 4.2.7.3.2

Combine the numerators over the common denominator.

$\frac{2 π \cdot 2 - π}{2}$

Step 4.2.7.4

Simplify the numerator.

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Step 4.2.7.4.1

Multiply $2$ by $2$ .

$\frac{4 π - π}{2}$

Step 4.2.7.4.2

Subtract $π$ from $4 π$ .

$\frac{3 π}{2}$

Step 4.2.7.5

List the new angles.

$x = \frac{3 π}{2}$

Step 4.2.8

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

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

Step 5

The final solution is all the values that make $(\sin (x) - 2) (\sin (x) + 1) = 0$ true.

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