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Popular >

pi/2 cos(pi/2 x)>= 0

Solution

\frac{π}{2} cos (\frac{π}{2} x) \geq 0

Solution

- 1 + 4 n \leq x \leq 1 + 4 n

Interval Notation

[- 1 + 4 n, 1 + 4 n]

Solution steps

\frac{π}{2} cos (\frac{π}{2} x) \geq 0

Multiply both sides by

2

\frac{π}{2} cos (\frac{π}{2} x) \geq 0

Multiply both sides by

2

2 \cdot \frac{π}{2} cos (\frac{π}{2} x) \geq 0 \cdot 2

Simplify

π cos (\frac{π}{2} x) \geq 0

π cos (\frac{π}{2} x) \geq 0

Divide both sides by

π

π cos (\frac{π}{2} x) \geq 0

Divide both sides by

π

\frac{π cos ( \frac{π}{2} x )}{π} \geq \frac{0}{π}

Simplify

cos (\frac{π}{2} x) \geq 0

cos (\frac{π}{2} x) \geq 0

For

cos (x) \geq a

, if

- 1 < a < 1

then

- arccos (a) + 2 πn \leq x \leq arccos (a) + 2 πn

- arccos (0) + 2 πn \leq \frac{π}{2} x \leq arccos (0) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (0) + 2 πn \leq \frac{π}{2} x and \frac{π}{2} x \leq arccos (0) + 2 πn

- arccos (0) + 2 πn \leq \frac{π}{2} x : x \geq - 1 + 4 n

- arccos (0) + 2 πn \leq \frac{π}{2} x

Switch sides

\frac{π}{2} x \geq - arccos (0) + 2 πn

Simplify

- arccos (0) + 2 πn : - \frac{π}{2} + 2 πn

- arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{π}{2} + 2 πn

\frac{π}{2} x \geq - \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{π}{2} x \geq - \frac{π}{2} + 2 πn

Multiply both sides by

2

2 \cdot \frac{π}{2} x \geq - 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

2 \cdot \frac{π}{2} x \geq - 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

2 \cdot \frac{π}{2} x : π x

2 \cdot \frac{π}{2} x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π}{2} x

Cancel the common factor:

2

= x π

Simplify

- 2 \cdot \frac{π}{2} + 2 \cdot 2 πn : - π + 4 πn

- 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= - π + 4 πn

π x \geq - π + 4 πn

π x \geq - π + 4 πn

π x \geq - π + 4 πn

Divide both sides by

π

π x \geq - π + 4 πn

Divide both sides by

π

\frac{π x}{π} \geq - \frac{π}{π} + \frac{4 πn}{π}

Simplify

\frac{π x}{π} \geq - \frac{π}{π} + \frac{4 πn}{π}

Simplify

\frac{π x}{π} : x

\frac{π x}{π}

Cancel the common factor:

π

= x

Simplify

- \frac{π}{π} + \frac{4 πn}{π} : - 1 + 4 n

- \frac{π}{π} + \frac{4 πn}{π}

Apply rule

\frac{a}{a} = 1

\frac{π}{π} = 1

= - 1 + \frac{4 πn}{π}

Cancel

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

Cancel the common factor:

π

= 4 n

= - 1 + 4 n

x \geq - 1 + 4 n

x \geq - 1 + 4 n

x \geq - 1 + 4 n

\frac{π}{2} x \leq arccos (0) + 2 πn : x \leq 1 + 4 n

\frac{π}{2} x \leq arccos (0) + 2 πn

Simplify

arccos (0) + 2 πn : \frac{π}{2} + 2 πn

arccos (0) + 2 πn

Use the following trivial identity:

arccos (0) = \frac{π}{2}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{2} + 2 πn

\frac{π}{2} x \leq \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{π}{2} x \leq \frac{π}{2} + 2 πn

Multiply both sides by

2

2 \cdot \frac{π}{2} x \leq 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

2 \cdot \frac{π}{2} x \leq 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

2 \cdot \frac{π}{2} x : π x

2 \cdot \frac{π}{2} x

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{2 π}{2} x

Cancel the common factor:

2

= x π

Simplify

2 \cdot \frac{π}{2} + 2 \cdot 2 πn : π + 4 πn

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

a \cdot \frac{b}{c} = \frac{a \cdot b}{c}

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

π x \leq π + 4 πn

π x \leq π + 4 πn

π x \leq π + 4 πn

Divide both sides by

π

π x \leq π + 4 πn

Divide both sides by

π

\frac{π x}{π} \leq \frac{π}{π} + \frac{4 πn}{π}

Simplify

\frac{π x}{π} \leq \frac{π}{π} + \frac{4 πn}{π}

Simplify

\frac{π x}{π} : x

\frac{π x}{π}

Cancel the common factor:

π

= x

Simplify

\frac{π}{π} + \frac{4 πn}{π} : 1 + 4 n

\frac{π}{π} + \frac{4 πn}{π}

Apply rule

\frac{a}{a} = 1

\frac{π}{π} = 1

= 1 + \frac{4 πn}{π}

Cancel

\frac{4 πn}{π} : 4 n

\frac{4 πn}{π}

Cancel the common factor:

π

= 4 n

= 1 + 4 n

x \leq 1 + 4 n

x \leq 1 + 4 n

x \leq 1 + 4 n

Combine the intervals

x \geq - 1 + 4 n and x \leq 1 + 4 n

Merge Overlapping Intervals

- 1 + 4 n \leq x \leq 1 + 4 n

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