Trigonometry Examples

$\sin (2 x) > \cos (2 x)$

Step 1

Divide each term in the equation by $\cos (2 x)$ .

$\frac{\sin (2 x)}{\cos (2 x)} > \frac{\cos (2 x)}{\cos (2 x)}$

Step 2

Convert from $\frac{\sin (2 x)}{\cos (2 x)}$ to $\tan (2 x)$ .

$\tan (2 x) > \frac{\cos (2 x)}{\cos (2 x)}$

Step 3

Cancel the common factor of $\cos (2 x)$ .

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Step 3.1

Cancel the common factor.

$\tan (2 x) > \frac{\cos (2 x)}{\cos (2 x)}$

Step 3.2

Rewrite the expression.

$\tan (2 x) > 1$

Step 4

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

$2 x > \arctan (1)$

Step 5

Simplify the right side.

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Step 5.1

The exact value of $\arctan (1)$ is $\frac{π}{4}$ .

$2 x > \frac{π}{4}$

Step 6

Divide each term in $2 x > \frac{π}{4}$ by $2$ and simplify.

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Step 6.1

Divide each term in $2 x > \frac{π}{4}$ by $2$ .

$\frac{2 x}{2} > \frac{\frac{π}{4}}{2}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $2$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{2 x}{2} > \frac{\frac{π}{4}}{2}$

Step 6.2.1.2

Divide $x$ by $1$ .

$x > \frac{\frac{π}{4}}{2}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Multiply the numerator by the reciprocal of the denominator.

$x > \frac{π}{4} \cdot \frac{1}{2}$

Step 6.3.2

Multiply $\frac{π}{4} \cdot \frac{1}{2}$ .

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Step 6.3.2.1

Multiply $\frac{π}{4}$ by $\frac{1}{2}$ .

$x > \frac{π}{4 \cdot 2}$

Step 6.3.2.2

Multiply $4$ by $2$ .

$x > \frac{π}{8}$

Step 7

The tangent function is positive in the first and third quadrants. To find the second solution, add the reference angle from $π$ to find the solution in the fourth quadrant.

$2 x = π + \frac{π}{4}$

Step 8

Solve for $x$ .

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Step 8.1

Simplify.

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Step 8.1.1

To write $π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

$2 x = π \cdot \frac{4}{4} + \frac{π}{4}$

Step 8.1.2

Combine $π$ and $\frac{4}{4}$ .

$2 x = \frac{π \cdot 4}{4} + \frac{π}{4}$

Step 8.1.3

Combine the numerators over the common denominator.

$2 x = \frac{π \cdot 4 + π}{4}$

Step 8.1.4

Add $π \cdot 4$ and $π$ .

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Step 8.1.4.1

Reorder $π$ and $4$ .

$2 x = \frac{4 \cdot π + π}{4}$

Step 8.1.4.2

Add $4 \cdot π$ and $π$ .

$2 x = \frac{5 \cdot π}{4}$

Step 8.2

Divide each term in $2 x = \frac{5 \cdot π}{4}$ by $2$ and simplify.

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Step 8.2.1

Divide each term in $2 x = \frac{5 \cdot π}{4}$ by $2$ .

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Step 8.2.2

Simplify the left side.

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Step 8.2.2.1

Cancel the common factor of $2$ .

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Step 8.2.2.1.1

Cancel the common factor.

$\frac{2 x}{2} = \frac{\frac{5 \cdot π}{4}}{2}$

Step 8.2.2.1.2

Divide $x$ by $1$ .

$x = \frac{\frac{5 \cdot π}{4}}{2}$

Step 8.2.3

Simplify the right side.

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Step 8.2.3.1

Multiply the numerator by the reciprocal of the denominator.

$x = \frac{5 \cdot π}{4} \cdot \frac{1}{2}$

Step 8.2.3.2

Multiply $\frac{5 π}{4} \cdot \frac{1}{2}$ .

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Step 8.2.3.2.1

Multiply $\frac{5 π}{4}$ by $\frac{1}{2}$ .

$x = \frac{5 π}{4 \cdot 2}$

Step 8.2.3.2.2

Multiply $4$ by $2$ .

$x = \frac{5 π}{8}$

Step 9

Find the period of $\tan (2 x)$ .

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Step 9.1

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

$\frac{π}{| b |}$

Step 9.2

Replace $b$ with $2$ in the formula for period.

$\frac{π}{| 2 |}$

Step 9.3

The absolute value is the distance between a number and zero. The distance between $0$ and $2$ is $2$ .

$\frac{π}{2}$

Step 10

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

$x = \frac{π}{8} + \frac{π n}{2}, \frac{5 π}{8} + \frac{π n}{2}$ , for any integer $n$

Step 11

Use each root to create test intervals.

$\frac{π}{8} < x < \frac{5 π}{8}$

$\frac{5 π}{8} < x < \frac{9 π}{8}$

Step 12

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 12.1

Test a value on the interval $\frac{π}{8} < x < \frac{5 π}{8}$ to see if it makes the inequality true.

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Step 12.1.1

Choose a value on the interval $\frac{π}{8} < x < \frac{5 π}{8}$ and see if this value makes the original inequality true.

$x = 1$

Step 12.1.2

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

$\sin (2 (1)) > \cos (2 (1))$

Step 12.1.3

The left side $0.90929742$ is greater than the right side $- 0.41614683$ , which means that the given statement is always true.

True

Step 12.2

Test a value on the interval $\frac{5 π}{8} < x < \frac{9 π}{8}$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $\frac{5 π}{8} < x < \frac{9 π}{8}$ and see if this value makes the original inequality true.

$x = 3$

Step 12.2.2

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

$\sin (2 (3)) > \cos (2 (3))$

Step 12.2.3

The left side $- 0.27941549$ is not greater than the right side $0.96017028$ , which means that the given statement is false.

False

Step 12.3

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

$\frac{π}{8} < x < \frac{5 π}{8}$ True

$\frac{5 π}{8} < x < \frac{9 π}{8}$ False

$\frac{π}{8} < x < \frac{5 π}{8}$ True

$\frac{5 π}{8} < x < \frac{9 π}{8}$ False

Step 13

The solution consists of all of the true intervals.

$\frac{π}{8} + \frac{π n}{2} < x < \frac{5 π}{8} + \frac{π n}{2}$ , for any integer $n$

Step 14