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Popular >

cos(2t)>= 1/(sqrt(2)),(0,2pi)

Solution

cos (2 t) \geq \frac{1}{2}, (0, 2 π)

Solution

0 < t \leq \frac{π}{8} or π - \frac{π}{8} \leq t \leq π + \frac{π}{8} or 2 π - \frac{π}{8} \leq t < 2 π

Interval Notation

(0, \frac{π}{8}] \cup [π - \frac{π}{8}, π + \frac{π}{8}] \cup [2 π - \frac{π}{8}, 2 π)

Decimal Notation

0 < t \leq 0.39269 \dots or 2.74889 \dots \leq t \leq 3.53429 \dots or 5.89048 \dots \leq t < 6.28318 \dots

Solution steps

cos (2 t) \geq \frac{1}{2}, 0 < t < 2 π

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

cos (2 t) \geq \frac{2}{2}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

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

a \leq u \leq b

then

a \leq u and u \leq b

- arccos (\frac{2}{2}) + 2 πn \leq 2 t and 2 t \leq arccos (\frac{2}{2}) + 2 πn

- arccos (\frac{2}{2}) + 2 πn \leq 2 t : t \geq - \frac{π}{8} + πn

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

Switch sides

2 t \geq - arccos (\frac{2}{2}) + 2 πn

Simplify

- arccos (\frac{2}{2}) + 2 πn : - \frac{π}{4} + 2 πn

- arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

2 t \geq - \frac{π}{4} + 2 πn

Divide both sides by

2

2 t \geq - \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 t}{2} \geq - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} \geq - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

- \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : - \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{π}{8} + πn

t \geq - \frac{π}{8} + πn

t \geq - \frac{π}{8} + πn

t \geq - \frac{π}{8} + πn

2 t \leq arccos (\frac{2}{2}) + 2 πn : t \leq \frac{π}{8} + πn

2 t \leq arccos (\frac{2}{2}) + 2 πn

Simplify

arccos (\frac{2}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{2}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{2}{2}) = \frac{π}{4}

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}{4} + 2 πn

2 t \leq \frac{π}{4} + 2 πn

Divide both sides by

2

2 t \leq \frac{π}{4} + 2 πn

Divide both sides by

2

\frac{2 t}{2} \leq \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} \leq \frac{\frac{π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

Simplify

\frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

t \leq \frac{π}{8} + πn

t \leq \frac{π}{8} + πn

t \leq \frac{π}{8} + πn

Combine the intervals

t \geq - \frac{π}{8} + πn and t \leq \frac{π}{8} + πn

Merge Overlapping Intervals

- \frac{π}{8} + πn \leq t \leq \frac{π}{8} + πn

Combine the intervals

- \frac{π}{8} + πn \leq t \leq \frac{π}{8} + πn and 0 < t < 2 π

0 < t \leq \frac{π}{8} or π - \frac{π}{8} \leq t \leq π + \frac{π}{8} or 2 π - \frac{π}{8} \leq t < 2 π

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