Precalculus Examples

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

Step 1

Move all the expressions to the left side of the equation.

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Step 1.1

Subtract $2 \cos (x)$ from both sides of the equation.

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

Step 1.2

Add $2$ to both sides of the equation.

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

Step 2

Replace $\sin^{2} (x)$ with $1 - \cos^{2} (x)$ .

$- (1 - \cos^{2} (x)) - 2 \cos (x) + 2 = 0$

Step 3

Solve for $x$ .

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Step 3.1

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

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

Step 3.2

Simplify $- (1 - u^{2}) - 2 u + 2$ .

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Apply the distributive property.

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

Step 3.2.1.2

Multiply $- 1$ by $1$ .

$- 1 - - u^{2} - 2 u + 2 = 0$

Step 3.2.1.3

Multiply $- - u^{2}$ .

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Step 3.2.1.3.1

Multiply $- 1$ by $- 1$ .

$- 1 + 1 u^{2} - 2 u + 2 = 0$

Step 3.2.1.3.2

Multiply $u^{2}$ by $1$ .

$- 1 + u^{2} - 2 u + 2 = 0$

Step 3.2.2

Add $- 1$ and $2$ .

$u^{2} - 2 u + 1 = 0$

Step 3.3

Factor using the perfect square rule.

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Step 3.3.1

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

$u^{2} - 2 u + 1^{2} = 0$

Step 3.3.2

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 3.3.3

Rewrite the polynomial.

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

Step 3.3.4

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

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

Step 3.4

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

$u - 1 = 0$

Step 3.5

Add $1$ to both sides of the equation.

$u = 1$

Step 3.6

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

$\cos (x) = 1$

Step 3.7

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

$x = \arccos (1)$

Step 3.8

Simplify the right side.

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Step 3.8.1

The exact value of $\arccos (1)$ is $0$ .

$x = 0$

Step 3.9

The cosine function is positive in the first and fourth quadrants. To find the second solution, subtract the reference angle from $2 π$ to find the solution in the fourth quadrant.

$x = 2 π - 0$

Step 3.10

Subtract $0$ from $2 π$ .

$x = 2 π$

Step 3.11

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

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Step 3.11.1

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

$\frac{2 π}{| b |}$

Step 3.11.2

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

$\frac{2 π}{| 1 |}$

Step 3.11.3

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

$\frac{2 π}{1}$

Step 3.11.4

Divide $2 π$ by $1$ .

$2 π$

Step 3.12

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

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

Step 4

Consolidate the answers.

$x = 2 π n$ , for any integer $n$