Trigonometry Examples

$2 \cos^{2} (3 x) + 5 \cos (3 x) - 3 < 0$

Step 1

Factor by grouping.

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Step 1.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 2 \cdot - 3 = - 6$ and whose sum is $b = 5$ .

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Step 1.1.1

Factor $5$ out of $5 \cos (3 x)$ .

$2 \cos^{2} (3 x) + 5 \cos (3 x) - 3 = 0$

Step 1.1.2

Rewrite $5$ as $- 1$ plus $6$

$2 \cos^{2} (3 x) + (- 1 + 6) \cos (3 x) - 3 = 0$

Step 1.1.3

Apply the distributive property.

$2 \cos^{2} (3 x) - 1 \cos (3 x) + 6 \cos (3 x) - 3 = 0$

Step 1.2

Factor out the greatest common factor from each group.

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Step 1.2.1

Group the first two terms and the last two terms.

$2 \cos^{2} (3 x) - 1 \cos (3 x) + 6 \cos (3 x) - 3 = 0$

Step 1.2.2

Factor out the greatest common factor (GCF) from each group.

$\cos (3 x) (2 \cos (3 x) - 1) + 3 (2 \cos (3 x) - 1) = 0$

Step 1.3

Factor the polynomial by factoring out the greatest common factor, $2 \cos (3 x) - 1$ .

$(2 \cos (3 x) - 1) (\cos (3 x) + 3) = 0$

Step 2

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

$2 \cos (3 x) - 1 = 0$

$\cos (3 x) + 3 = 0$

Step 3

Set $2 \cos (3 x) - 1$ equal to $0$ and solve for $x$ .

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Step 3.1

Set $2 \cos (3 x) - 1$ equal to $0$ .

$2 \cos (3 x) - 1 = 0$

Step 3.2

Solve $2 \cos (3 x) - 1 = 0$ for $x$ .

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Step 3.2.1

Add $1$ to both sides of the equation.

$2 \cos (3 x) = 1$

Step 3.2.2

Divide each term in $2 \cos (3 x) = 1$ by $2$ and simplify.

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Step 3.2.2.1

Divide each term in $2 \cos (3 x) = 1$ by $2$ .

$\frac{2 \cos (3 x)}{2} = \frac{1}{2}$

Step 3.2.2.2

Simplify the left side.

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Step 3.2.2.2.1

Cancel the common factor of $2$ .

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Step 3.2.2.2.1.1

Cancel the common factor.

$\frac{2 \cos (3 x)}{2} = \frac{1}{2}$

Step 3.2.2.2.1.2

Divide $\cos (3 x)$ by $1$ .

$\cos (3 x) = \frac{1}{2}$

Step 3.2.3

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

$3 x = \arccos (\frac{1}{2})$

Step 3.2.4

Simplify the right side.

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Step 3.2.4.1

The exact value of $\arccos (\frac{1}{2})$ is $\frac{π}{3}$ .

$3 x = \frac{π}{3}$

Step 3.2.5

Divide each term in $3 x = \frac{π}{3}$ by $3$ and simplify.

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Step 3.2.5.1

Divide each term in $3 x = \frac{π}{3}$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{π}{3}}{3}$

Step 3.2.5.2

Simplify the left side.

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Step 3.2.5.2.1

Cancel the common factor of $3$ .

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Step 3.2.5.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{π}{3}}{3}$

Step 3.2.5.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{π}{3}}{3}$

Step 3.2.5.3

Simplify the right side.

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Step 3.2.5.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{π}{3} \cdot \frac{1}{3}$

Step 3.2.5.3.2

Multiply $\frac{π}{3} \cdot \frac{1}{3}$ .

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Step 3.2.5.3.2.1

Multiply $\frac{π}{3}$ by $\frac{1}{3}$ .

$x = \frac{π}{3 \cdot 3}$

Step 3.2.5.3.2.2

Multiply $3$ by $3$ .

$x = \frac{π}{9}$

Step 3.2.6

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.

$3 x = 2 π - \frac{π}{3}$

Step 3.2.7

Solve for $x$ .

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Step 3.2.7.1

Simplify.

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Step 3.2.7.1.1

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{3}{3}$ .

$3 x = 2 π \cdot \frac{3}{3} - \frac{π}{3}$

Step 3.2.7.1.2

Combine $2 π$ and $\frac{3}{3}$ .

$3 x = \frac{2 π \cdot 3}{3} - \frac{π}{3}$

Step 3.2.7.1.3

Combine the numerators over the common denominator.

$3 x = \frac{2 π \cdot 3 - π}{3}$

Step 3.2.7.1.4

Multiply $3$ by $2$ .

$3 x = \frac{6 π - π}{3}$

Step 3.2.7.1.5

Subtract $π$ from $6 π$ .

$3 x = \frac{5 π}{3}$

Step 3.2.7.2

Divide each term in $3 x = \frac{5 π}{3}$ by $3$ and simplify.

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Step 3.2.7.2.1

Divide each term in $3 x = \frac{5 π}{3}$ by $3$ .

$\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3}$

Step 3.2.7.2.2

Simplify the left side.

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Step 3.2.7.2.2.1

Cancel the common factor of $3$ .

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Step 3.2.7.2.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{\frac{5 π}{3}}{3}$

Step 3.2.7.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 π}{3}}{3}$

Step 3.2.7.2.3

Simplify the right side.

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Step 3.2.7.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 π}{3} \cdot \frac{1}{3}$

Step 3.2.7.2.3.2

Multiply $\frac{5 π}{3} \cdot \frac{1}{3}$ .

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Step 3.2.7.2.3.2.1

Multiply $\frac{5 π}{3}$ by $\frac{1}{3}$ .

$x = \frac{5 π}{3 \cdot 3}$

Step 3.2.7.2.3.2.2

Multiply $3$ by $3$ .

$x = \frac{5 π}{9}$

Step 3.2.8

Find the period of $\cos (3 x)$ .

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Step 3.2.8.1

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

$\frac{2 π}{| b |}$

Step 3.2.8.2

Replace $b$ with $3$ in the formula for period.

$\frac{2 π}{| 3 |}$

Step 3.2.8.3

The absolute value is the distance between a number and zero. The distance between $0$ and $3$ is $3$ .

$\frac{2 π}{3}$

Step 3.2.9

The period of the $\cos (3 x)$ function is $\frac{2 π}{3}$ so values will repeat every $\frac{2 π}{3}$ radians in both directions.

$x = \frac{π}{9} + \frac{2 π n}{3}, \frac{5 π}{9} + \frac{2 π n}{3}$ , for any integer $n$

Step 4

Set $\cos (3 x) + 3$ equal to $0$ and solve for $x$ .

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Step 4.1

Set $\cos (3 x) + 3$ equal to $0$ .

$\cos (3 x) + 3 = 0$

Step 4.2

Solve $\cos (3 x) + 3 = 0$ for $x$ .

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Step 4.2.1

Subtract $3$ from both sides of the equation.

$\cos (3 x) = - 3$

Step 4.2.2

The range of cosine is $- 1 \leq y \leq 1$ . Since $- 3$ does not fall in this range, there is no solution.

No solution

Step 5

The final solution is all the values that make $(2 \cos (3 x) - 1) (\cos (3 x) + 3) = 0$ true.

$x = \frac{π}{9} + \frac{2 π n}{3}, \frac{5 π}{9} + \frac{2 π n}{3}$ , for any integer $n$

Step 6

Use each root to create test intervals.

$\frac{π}{9} < x < \frac{5 π}{9}$

$\frac{5 π}{9} < x < \frac{7 π}{9}$

$\frac{7 π}{9} < x < \frac{11 π}{9}$

Step 7

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 7.1

Test a value on the interval $\frac{π}{9} < x < \frac{5 π}{9}$ to see if it makes the inequality true.

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Step 7.1.1

Choose a value on the interval $\frac{π}{9} < x < \frac{5 π}{9}$ and see if this value makes the original inequality true.

$x = 1$

Step 7.1.2

Replace $x$ with $1$ in the original inequality.

$2 \cos^{2} (3 (1)) + 5 \cos (3 (1)) - 3 < 0$

Step 7.1.3

The left side $- 5.98979219$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 7.2

Test a value on the interval $\frac{5 π}{9} < x < \frac{7 π}{9}$ to see if it makes the inequality true.

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Step 7.2.1

Choose a value on the interval $\frac{5 π}{9} < x < \frac{7 π}{9}$ and see if this value makes the original inequality true.

$x = 2$

Step 7.2.2

Replace $x$ with $2$ in the original inequality.

$2 \cos^{2} (3 (2)) + 5 \cos (3 (2)) - 3 < 0$

Step 7.2.3

The left side $3.64470539$ is not less than the right side $0$ , which means that the given statement is false.

False

Step 7.3

Test a value on the interval $\frac{7 π}{9} < x < \frac{11 π}{9}$ to see if it makes the inequality true.

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Step 7.3.1

Choose a value on the interval $\frac{7 π}{9} < x < \frac{11 π}{9}$ and see if this value makes the original inequality true.

$x = 3$

Step 7.3.2

Replace $x$ with $3$ in the original inequality.

$2 \cos^{2} (3 (3)) + 5 \cos (3 (3)) - 3 < 0$

Step 7.3.3

The left side $- 5.8953346$ is less than the right side $0$ , which means that the given statement is always true.

True

Step 7.4

Compare the intervals to determine which ones satisfy the original inequality.

$\frac{π}{9} < x < \frac{5 π}{9}$ True

$\frac{5 π}{9} < x < \frac{7 π}{9}$ False

$\frac{7 π}{9} < x < \frac{11 π}{9}$ True

$\frac{π}{9} < x < \frac{5 π}{9}$ True

$\frac{5 π}{9} < x < \frac{7 π}{9}$ False

$\frac{7 π}{9} < x < \frac{11 π}{9}$ True

Step 8

The solution consists of all of the true intervals.

$\frac{π}{9} + \frac{2 π n}{3} < x < \frac{5 π}{9} + \frac{2 π n}{3}$ or $\frac{7 π}{9} + \frac{2 π n}{3} < x < \frac{11 π}{9} + \frac{2 π n}{3}$ , for any integer $n$

Step 9