Solve the Inequality for x cos(x)>=( square root of 2)/2

$\cos (x) \geq \frac{\sqrt{2}}{2}$

Step 1

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

$x \geq \arccos (\frac{\sqrt{2}}{2})$

Step 2

Simplify the right side.

The exact value of $\arccos (\frac{\sqrt{2}}{2})$ is $\frac{π}{4}$ .

$x \geq \frac{π}{4}$

Step 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.

$x = 2 π - \frac{π}{4}$

Step 4

Simplify $2 π - \frac{π}{4}$ .

To write $2 π$ as a fraction with a common denominator, multiply by $\frac{4}{4}$ .

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

Combine fractions.

Combine $2 π$ and $\frac{4}{4}$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 4 - π}{4}$

Simplify the numerator.

Multiply $4$ by $2$ .

$x = \frac{8 π - π}{4}$

Subtract $π$ from $8 π$ .

$x = \frac{7 π}{4}$

Step 5

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

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

$\frac{2 π}{| b |}$

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

$\frac{2 π}{| 1 |}$

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

$\frac{2 π}{1}$

Divide $2 π$ by $1$ .

$2 π$

Step 6

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

$x = \frac{π}{4} + 2 π n, \frac{7 π}{4} + 2 π n$ , for any integer $n$

Step 7

Use each root to create test intervals.

$\frac{π}{4} < x < \frac{7 π}{4}$

$\frac{7 π}{4} < x < \frac{9 π}{4}$

$\frac{9 π}{4} < x < \frac{15 π}{4}$

Step 8

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

Test a value on the interval $\frac{π}{4} < x < \frac{7 π}{4}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{π}{4} < x < \frac{7 π}{4}$ and see if this value makes the original inequality true.

$x = 3$

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

$\cos (3) \geq \frac{\sqrt{2}}{2}$

The left side $- 0.98999249$ is less than the right side $0.70710678$ , which means that the given statement is false.

False

Test a value on the interval $\frac{7 π}{4} < x < \frac{9 π}{4}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{7 π}{4} < x < \frac{9 π}{4}$ and see if this value makes the original inequality true.

$x = 6$

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

$\cos (6) \geq \frac{\sqrt{2}}{2}$

The left side $0.96017028$ is greater than the right side $0.70710678$ , which means that the given statement is always true.

True

Test a value on the interval $\frac{9 π}{4} < x < \frac{15 π}{4}$ to see if it makes the inequality true.

Choose a value on the interval $\frac{9 π}{4} < x < \frac{15 π}{4}$ and see if this value makes the original inequality true.

$x = 9$

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

$\cos (9) \geq \frac{\sqrt{2}}{2}$

The left side $- 0.91113026$ is less than the right side $0.70710678$ , which means that the given statement is false.

False

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

$\frac{π}{4} < x < \frac{7 π}{4}$ False

$\frac{7 π}{4} < x < \frac{9 π}{4}$ True

$\frac{9 π}{4} < x < \frac{15 π}{4}$ False

$\frac{π}{4} < x < \frac{7 π}{4}$ False

$\frac{7 π}{4} < x < \frac{9 π}{4}$ True

$\frac{9 π}{4} < x < \frac{15 π}{4}$ False

Step 9

The solution consists of all of the true intervals.

$\frac{7 π}{4} + 2 π n \leq x \leq \frac{9 π}{4} + 2 π n$ , for any integer $n$

Step 10

Solve the Inequality for x cos(x)>=( square root of 2)/2

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