Trigonometry Examples

$2 \sin^{2} (x) > 3 \cos (x)$

Step 1

Subtract $3 \cos (x)$ from both sides of the inequality.

$2 \sin^{2} (x) - 3 \cos (x) > 0$

Step 2

Replace the $2 \sin^{2} (x)$ with $2 (1 - \cos^{2} (x))$ based on the $\sin^{2} (x) + \cos^{2} (x) = 1$ identity.

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

Step 3

Simplify each term.

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Step 3.1

Apply the distributive property.

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

Step 3.2

Multiply $2$ by $1$ .

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

Step 3.3

Multiply $- 1$ by $2$ .

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

Step 4

Reorder the polynomial.

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

Step 5

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

$- 2 {(u)}^{2} - 3 u + 2 = 0$

Step 6

Factor the left side of the equation.

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Step 6.1

Factor $- 1$ out of $- 2 u^{2} - 3 u + 2$ .

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Step 6.1.1

Factor $- 1$ out of $- 2 u^{2}$ .

$- (2 u^{2}) - 3 u + 2 = 0$

Step 6.1.2

Factor $- 1$ out of $- 3 u$ .

$- (2 u^{2}) - (3 u) + 2 = 0$

Step 6.1.3

Rewrite $2$ as $- 1 (- 2)$ .

$- (2 u^{2}) - (3 u) - 1 \cdot - 2 = 0$

Step 6.1.4

Factor $- 1$ out of $- (2 u^{2}) - (3 u)$ .

$- (2 u^{2} + 3 u) - 1 \cdot - 2 = 0$

Step 6.1.5

Factor $- 1$ out of $- (2 u^{2} + 3 u) - 1 (- 2)$ .

$- (2 u^{2} + 3 u - 2) = 0$

Step 6.2

Factor.

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Step 6.2.1

Factor by grouping.

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Step 6.2.1.1

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 $a \cdot c = 2 \cdot - 2 = - 4$ and whose sum is $b = 3$ .

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Step 6.2.1.1.1

Factor $3$ out of $3 u$ .

$- (2 u^{2} + 3 u - 2) = 0$

Step 6.2.1.1.2

Rewrite $3$ as $- 1$ plus $4$

$- (2 u^{2} + (- 1 + 4) u - 2) = 0$

Step 6.2.1.1.3

Apply the distributive property.

$- (2 u^{2} - 1 u + 4 u - 2) = 0$

Step 6.2.1.2

Factor out the greatest common factor from each group.

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Step 6.2.1.2.1

Group the first two terms and the last two terms.

$- (2 u^{2} - 1 u + 4 u - 2) = 0$

Step 6.2.1.2.2

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

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

Step 6.2.1.3

Factor the polynomial by factoring out the greatest common factor, $2 u - 1$ .

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

Step 6.2.2

Remove unnecessary parentheses.

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

Step 7

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

$2 u - 1 = 0$

$u + 2 = 0$

Step 8

Set $2 u - 1$ equal to $0$ and solve for $u$ .

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Step 8.1

Set $2 u - 1$ equal to $0$ .

$2 u - 1 = 0$

Step 8.2

Solve $2 u - 1 = 0$ for $u$ .

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Step 8.2.1

Add $1$ to both sides of the equation.

$2 u = 1$

Step 8.2.2

Divide each term in $2 u = 1$ by $2$ and simplify.

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Step 8.2.2.1

Divide each term in $2 u = 1$ by $2$ .

$\frac{2 u}{2} = \frac{1}{2}$

Step 8.2.2.2

Simplify the left side.

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Step 8.2.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.2.1.1

Cancel the common factor.

$\frac{2 u}{2} = \frac{1}{2}$

Step 8.2.2.2.1.2

Divide $u$ by $1$ .

$u = \frac{1}{2}$

Step 9

Set $u + 2$ equal to $0$ and solve for $u$ .

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Step 9.1

Set $u + 2$ equal to $0$ .

$u + 2 = 0$

Step 9.2

Subtract $2$ from both sides of the equation.

$u = - 2$

Step 10

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

$u = \frac{1}{2}, - 2$

Step 11

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

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

Step 12

Set up each of the solutions to solve for $x$ .

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

$\cos (x) = - 2$

Step 13

Solve for $x$ in $\cos (x) = \frac{1}{2}$ .

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Step 13.1

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

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

Step 13.2

Simplify the right side.

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Step 13.2.1

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

$x = \frac{π}{3}$

Step 13.3

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.

$x = 2 π - \frac{π}{3}$

Step 13.4

Simplify $2 π - \frac{π}{3}$ .

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Step 13.4.1

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

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

Step 13.4.2

Combine fractions.

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Step 13.4.2.1

Combine $2 π$ and $\frac{3}{3}$ .

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

Step 13.4.2.2

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Step 13.4.3

Simplify the numerator.

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Step 13.4.3.1

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Step 13.4.3.2

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

Step 13.5

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

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Step 13.5.1

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

$\frac{2 π}{| b |}$

Step 13.5.2

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

$\frac{2 π}{| 1 |}$

Step 13.5.3

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

$\frac{2 π}{1}$

Step 13.5.4

Divide $2 π$ by $1$ .

$2 π$

Step 13.6

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

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

Step 14

Solve for $x$ in $\cos (x) = - 2$ .

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Step 14.1

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

No solution

Step 15

List all of the solutions.

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

Step 16

Use each root to create test intervals.

$\frac{π}{3} < x < \frac{5 π}{3}$

$\frac{5 π}{3} < x < \frac{7 π}{3}$

$\frac{7 π}{3} < x < \frac{11 π}{3}$

Step 17

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 17.1

Test a value on the interval $\frac{π}{3} < x < \frac{5 π}{3}$ to see if it makes the inequality true.

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Step 17.1.1

Choose a value on the interval $\frac{π}{3} < x < \frac{5 π}{3}$ and see if this value makes the original inequality true.

$x = 3$

Step 17.1.2

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

$2 \sin^{2} (3) > 3 \cos (3)$

Step 17.1.3

The left side $0.03982971$ is greater than the right side $- 2.96997748$ , which means that the given statement is always true.

True

Step 17.2

Test a value on the interval $\frac{5 π}{3} < x < \frac{7 π}{3}$ to see if it makes the inequality true.

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Step 17.2.1

Choose a value on the interval $\frac{5 π}{3} < x < \frac{7 π}{3}$ and see if this value makes the original inequality true.

$x = 6$

Step 17.2.2

Replace $x$ with $6$ in the original inequality.

$2 \sin^{2} (6) > 3 \cos (6)$

Step 17.2.3

The left side $0.15614604$ is not greater than the right side $2.88051085$ , which means that the given statement is false.

False

Step 17.3

Test a value on the interval $\frac{7 π}{3} < x < \frac{11 π}{3}$ to see if it makes the inequality true.

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Step 17.3.1

Choose a value on the interval $\frac{7 π}{3} < x < \frac{11 π}{3}$ and see if this value makes the original inequality true.

$x = 9$

Step 17.3.2

Replace $x$ with $9$ in the original inequality.

$2 \sin^{2} (9) > 3 \cos (9)$

Step 17.3.3

The left side $0.33968329$ is greater than the right side $- 2.73339078$ , which means that the given statement is always true.

True

Step 17.4

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

$\frac{π}{3} < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < \frac{7 π}{3}$ False

$\frac{7 π}{3} < x < \frac{11 π}{3}$ True

$\frac{π}{3} < x < \frac{5 π}{3}$ True

$\frac{5 π}{3} < x < \frac{7 π}{3}$ False

$\frac{7 π}{3} < x < \frac{11 π}{3}$ True

Step 18

The solution consists of all of the true intervals.

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

Step 19