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Popular >

2sin(2x)+1/2 >= 0

Solution

2 sin (2 x) + \frac{1}{2} \geq 0

Solution

- \frac{arcsin ( \frac{1}{4} )}{2} + πn \leq x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

+2

Interval Notation

[- \frac{arcsin ( \frac{1}{4} )}{2} + πn, \frac{π + arcsin ( \frac{1}{4} )}{2} + πn]

Decimal Notation

- 0.12634 \dots + πn \leq x \leq 1.69713 \dots + πn

Solution steps

2 sin (2 x) + \frac{1}{2} \geq 0

Move

\frac{1}{2}

to the right side

2 sin (2 x) + \frac{1}{2} \geq 0

Subtract

\frac{1}{2}

from both sides

2 sin (2 x) + \frac{1}{2} - \frac{1}{2} \geq 0 - \frac{1}{2}

Simplify

2 sin (2 x) \geq - \frac{1}{2}

2 sin (2 x) \geq - \frac{1}{2}

Divide both sides by

2

2 sin (2 x) \geq - \frac{1}{2}

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} \geq \frac{- \frac{1}{2}}{2}

Simplify

\frac{2 sin ( 2 x )}{2} \geq \frac{- \frac{1}{2}}{2}

Simplify

\frac{2 sin ( 2 x )}{2} : sin (2 x)

\frac{2 sin ( 2 x )}{2}

Divide the numbers:

\frac{2}{2} = 1

= sin (2 x)

Simplify

\frac{- \frac{1}{2}}{2} : - \frac{1}{4}

\frac{- \frac{1}{2}}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{\frac{1}{2}}{2}

Apply the fraction rule:

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

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

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

Multiply the numbers:

2 \cdot 2 = 4

= - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

sin (2 x) \geq - \frac{1}{4}

For

sin (x) \geq a

, if

- 1 < a < 1

then

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

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

If

a \leq u \leq b

then

a \leq u and u \leq b

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x and 2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x : x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

arcsin (- \frac{1}{4}) + 2 πn \leq 2 x

Switch sides

2 x \geq arcsin (- \frac{1}{4}) + 2 πn

Simplify

arcsin (- \frac{1}{4}) + 2 πn : - arcsin (\frac{1}{4}) + 2 πn

arcsin (- \frac{1}{4}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{4}) = - arcsin (\frac{1}{4})

= - arcsin (\frac{1}{4}) + 2 πn

2 x \geq - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x \geq - arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} \geq - \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplify

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn

2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn : x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

2 x \leq π - arcsin (- \frac{1}{4}) + 2 πn

Simplify

π - arcsin (- \frac{1}{4}) + 2 πn : π + arcsin (\frac{1}{4}) + 2 πn

π - arcsin (- \frac{1}{4}) + 2 πn

Use the following property:

arcsin (- x) = - arcsin (x)

arcsin (- \frac{1}{4}) = - arcsin (\frac{1}{4})

= π - (- arcsin (\frac{1}{4})) + 2 πn

Apply rule

- (- a) = a

= π + arcsin (\frac{1}{4}) + 2 πn

2 x \leq π + arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

2 x \leq π + arcsin (\frac{1}{4}) + 2 πn

Divide both sides by

2

\frac{2 x}{2} \leq \frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} + \frac{2 πn}{2}

Simplify

x \leq \frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} + πn

Simplify

\frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2} : \frac{π + arcsin ( \frac{1}{4} )}{2}

\frac{π}{2} + \frac{arcsin ( \frac{1}{4} )}{2}

Apply rule

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

= \frac{π + arcsin ( \frac{1}{4} )}{2}

x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

Combine the intervals

x \geq - \frac{arcsin ( \frac{1}{4} )}{2} + πn and x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

Merge Overlapping Intervals

- \frac{arcsin ( \frac{1}{4} )}{2} + πn \leq x \leq \frac{π + arcsin ( \frac{1}{4} )}{2} + πn

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