Precalculus Examples

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

Step 1

Set $- \cos (2 x) + \cos^{2} (x)$ equal to $0$ .

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

Step 2

Solve for $x$ .

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Step 2.1

Use the double-angle identity to transform $\cos (2 x)$ to $1 - 2 \sin^{2} (x)$ .

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $- (1 - 2 \sin^{2} (x)) + \cos^{2} (x)$ .

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Step 2.2.1.1

Simplify terms.

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Step 2.2.1.1.1

Simplify each term.

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Step 2.2.1.1.1.1

Apply the distributive property.

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

Step 2.2.1.1.1.2

Multiply $- 1$ by $1$ .

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

Step 2.2.1.1.1.3

Multiply $- 2$ by $- 1$ .

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

Step 2.2.1.1.2

Simplify with factoring out.

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Step 2.2.1.1.2.1

Move $\cos^{2} (x)$ .

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

Step 2.2.1.1.2.2

Rewrite $- 1$ as $- 1 (1)$ .

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

Step 2.2.1.1.2.3

Factor $- 1$ out of $\cos^{2} (x)$ .

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

Step 2.2.1.1.2.4

Factor $- 1$ out of $- 1 (1) - 1 (- \cos^{2} (x))$ .

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

Step 2.2.1.1.2.5

Rewrite $- 1 (1 - \cos^{2} (x))$ as $- (1 - \cos^{2} (x))$ .

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

Step 2.2.1.2

Apply pythagorean identity.

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

Step 2.2.1.3

Add $- \sin^{2} (x)$ and $2 \sin^{2} (x)$ .

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

Step 2.3

Solve the equation for $x$ .

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Step 2.3.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$\sin (x) = \pm \sqrt{0}$

Step 2.3.2

Simplify $\pm \sqrt{0}$ .

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Step 2.3.2.1

Rewrite $0$ as $0^{2}$ .

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

Step 2.3.2.2

Pull terms out from under the radical, assuming positive real numbers.

$\sin (x) = \pm 0$

Step 2.3.2.3

Plus or minus $0$ is $0$ .

$\sin (x) = 0$

Step 2.3.3

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

$x = \arcsin (0)$

Step 2.3.4

Simplify the right side.

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Step 2.3.4.1

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

$x = 0$

Step 2.3.5

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.

$x = π - 0$

Step 2.3.6

Subtract $0$ from $π$ .

$x = π$

Step 2.3.7

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

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Step 2.3.7.1

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

$\frac{2 π}{| b |}$

Step 2.3.7.2

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

$\frac{2 π}{| 1 |}$

Step 2.3.7.3

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

$\frac{2 π}{1}$

Step 2.3.7.4

Divide $2 π$ by $1$ .

$2 π$

Step 2.3.8

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

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

Step 2.4

Consolidate the answers.

$x = π n$ , for any integer $n$

Step 3