Solve the Inequality for x cos(x)>=1/2

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

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

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

Simplify the right side.

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

$x \geq \frac{π}{3}$

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.

$x = 2 π - \frac{π}{3}$

Simplify $2 π - \frac{π}{3}$ .

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

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

Combine fractions.

Combine $2 π$ and $\frac{3}{3}$ .

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

Combine the numerators over the common denominator.

$x = \frac{2 π \cdot 3 - π}{3}$

Simplify the numerator.

Multiply $3$ by $2$ .

$x = \frac{6 π - π}{3}$

Subtract $π$ from $6 π$ .

$x = \frac{5 π}{3}$

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 $1$ .

$2 π$

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 $n$

Use each root to create test intervals.

$\frac{π}{3} < x < \frac{5 π}{3}$

$\frac{5 π}{3} < x < \frac{7 π}{3}$

$\frac{7 π}{3} < x < \frac{11 π}{3}$

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 $\frac{π}{3} < x < \frac{5 π}{3}$ to see if it makes the inequality true.

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

$x = 3$

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

$\cos (3) \geq \frac{1}{2}$

The left side $- 0.98999249$ is less than the right side $0.5$ , which means that the given statement is false.

False

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

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.

$x = 6$

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

$\cos (6) \geq \frac{1}{2}$

The left side $0.96017028$ is greater than the right side $0.5$ , which means that the given statement is always true.

True

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

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.

$x = 9$

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

$\cos (9) \geq \frac{1}{2}$

The left side $- 0.91113026$ is less than the right side $0.5$ , which means that the given statement is false.

False

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

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

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

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

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

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

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

The solution consists of all of the true intervals.

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

Solve the Inequality for x cos(x)>=1/2

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