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Popular >

2cos(x)-1>= 0

Solution

2 cos (x) - 1 \geq 0

Solution

- \frac{π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

+2

Interval Notation

[- \frac{π}{3} + 2 πn, \frac{π}{3} + 2 πn]

Decimal Notation

- 1.04719 \dots + 2 πn \leq x \leq 1.04719 \dots + 2 πn

Solution steps

2 cos (x) - 1 \geq 0

Move

1

to the right side

2 cos (x) - 1 \geq 0

Add

1

to both sides

2 cos (x) - 1 + 1 \geq 0 + 1

Simplify

2 cos (x) \geq 1

2 cos (x) \geq 1

Divide both sides by

2

2 cos (x) \geq 1

Divide both sides by

2

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

Simplify

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

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

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

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

Simplify

- arccos (\frac{1}{2}) : - \frac{π}{3}

- arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

Simplify

arccos (\frac{1}{2}) : \frac{π}{3}

arccos (\frac{1}{2})

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{3}

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}{3}

- \frac{π}{3} + 2 πn \leq x \leq \frac{π}{3} + 2 πn

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