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Popular >

-1>=-cos(2x)>= 1

Solution

- 1 \geq - cos (2 x) \geq 1

Solution

False for all x \in R

Solution steps

- 1 \geq - cos (2 x) \geq 1

a \geq u \geq b

then

a \geq u and u \geq b

- 1 \geq - cos (2 x) and - cos (2 x) \geq 1

- 1 \geq - cos (2 x) :

No Solution for

x \in R

- 1 \geq - cos (2 x)

Switch sides

- cos (2 x) \leq - 1

Multiply both sides by

- 1

- cos (2 x) \leq - 1

Multiply both sides by -1 (reverse the inequality)

(- cos (2 x)) (- 1) \geq (- 1) (- 1)

Simplify

cos (2 x) \geq 1

cos (2 x) \geq 1

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (1) + 2 πn \leq 2 x \leq arccos (1) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (1) + 2 πn \leq 2 x and 2 x \leq arccos (1) + 2 πn

- arccos (1) + 2 πn \leq 2 x : x \geq πn

- arccos (1) + 2 πn \leq 2 x

Switch sides

2 x \geq - arccos (1) + 2 πn

Simplify

- arccos (1) + 2 πn : 2 πn

- arccos (1) + 2 πn

Use the following trivial identity:

arccos (1) = 0

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}

= - 0 + 2 πn

- 0 + 2 πn = 2 πn

= 2 πn

2 x \geq 2 πn

Divide both sides by

2

2 x \geq 2 πn

Divide both sides by

2

\frac{2 x}{2} \geq \frac{2 πn}{2}

Simplify

x \geq πn

x \geq πn

2 x \leq arccos (1) + 2 πn : x \leq πn

2 x \leq arccos (1) + 2 πn

Simplify

arccos (1) + 2 πn : 2 πn

arccos (1) + 2 πn

Use the following trivial identity:

arccos (1) = 0

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}

= 0 + 2 πn

0 + 2 πn = 2 πn

= 2 πn

2 x \leq 2 πn

Divide both sides by

2

2 x \leq 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{2 πn}{2}

Simplify

x \leq πn

x \leq πn

Combine the intervals

x \geq πn and x \leq πn

Merge Overlapping Intervals

No Solution for x \in R

- cos (2 x) \geq 1 : x = \frac{π}{2} + πn

- cos (2 x) \geq 1

Multiply both sides by

- 1

- cos (2 x) \geq 1

Multiply both sides by -1 (reverse the inequality)

(- cos (2 x)) (- 1) \leq 1 \cdot (- 1)

Simplify

cos (2 x) \leq - 1

cos (2 x) \leq - 1

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (- 1) + 2 πn \leq 2 x \leq 2 π - arccos (- 1) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arccos (- 1) + 2 πn \leq 2 x and 2 x \leq 2 π - arccos (- 1) + 2 πn

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

arccos (- 1) + 2 πn \leq 2 x

Switch sides

2 x \geq arccos (- 1) + 2 πn

Simplify

arccos (- 1) + 2 πn : π + 2 πn

arccos (- 1) + 2 πn

Use the following trivial identity:

arccos (- 1) = π

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}

= π + 2 πn

2 x \geq π + 2 πn

Divide both sides by

2

2 x \geq π + 2 πn

Divide both sides by

2

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

Simplify

x \geq \frac{π}{2} + πn

x \geq \frac{π}{2} + πn

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

2 x \leq 2 π - arccos (- 1) + 2 πn

Simplify

2 π - arccos (- 1) + 2 πn : π + 2 πn

2 π - arccos (- 1) + 2 πn

Use the following trivial identity:

arccos (- 1) = π

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}

= 2 π - π + 2 πn

Add similar elements:

2 π - π = π

= π + 2 πn

2 x \leq π + 2 πn

Divide both sides by

2

2 x \leq π + 2 πn

Divide both sides by

2

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

Simplify

x \leq \frac{π}{2} + πn

x \leq \frac{π}{2} + πn

Combine the intervals

x \geq \frac{π}{2} + πn and x \leq \frac{π}{2} + πn

Merge Overlapping Intervals

x = \frac{π}{2} + πn

Combine the intervals

False for all x \in R and x = \frac{π}{2} + πn

Merge Overlapping Intervals

False for all x \in R

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