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4cos(pi/2+x)+3<= 0

Solution

4 cos (\frac{π}{2} + x) + 3 \leq 0

Solution

\frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn \leq x \leq \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn

Interval Notation

[\frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn, \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn]

Decimal

0.84806 \dots + 2 πn \leq x \leq 2.29353 \dots + 2 πn

Solution steps

4 cos (\frac{π}{2} + x) + 3 \leq 0

Move

3

to the right side

4 cos (\frac{π}{2} + x) + 3 \leq 0

Subtract

3

from both sides

4 cos (\frac{π}{2} + x) + 3 - 3 \leq 0 - 3

Simplify

4 cos (\frac{π}{2} + x) \leq - 3

4 cos (\frac{π}{2} + x) \leq - 3

Divide both sides by

4

4 cos (\frac{π}{2} + x) \leq - 3

Divide both sides by

4

\frac{4 cos ( \frac{π}{2} + x )}{4} \leq \frac{- 3}{4}

Simplify

cos (\frac{π}{2} + x) \leq - \frac{3}{4}

cos (\frac{π}{2} + x) \leq - \frac{3}{4}

For

cos (x) \leq a

, if

- 1 < a < 1

then

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

arccos (- \frac{3}{4}) + 2 πn \leq (\frac{π}{2} + x) \leq 2 π - arccos (- \frac{3}{4}) + 2 πn

a \leq u \leq b

then

a \leq u and u \leq b

arccos (- \frac{3}{4}) + 2 πn \leq \frac{π}{2} + x and \frac{π}{2} + x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn

arccos (- \frac{3}{4}) + 2 πn \leq \frac{π}{2} + x : x \geq \frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn

arccos (- \frac{3}{4}) + 2 πn \leq \frac{π}{2} + x

Switch sides

\frac{π}{2} + x \geq arccos (- \frac{3}{4}) + 2 πn

Move

\frac{π}{2}

to the right side

\frac{π}{2} + x \geq arccos (- \frac{3}{4}) + 2 πn

Subtract

\frac{π}{2}

from both sides

\frac{π}{2} + x - \frac{π}{2} \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

Simplify

x \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

x \geq arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

Simplify

arccos (- \frac{3}{4}) - \frac{π}{2} : \frac{2 arccos ( - \frac{3}{4} ) - π}{2}

arccos (- \frac{3}{4}) - \frac{π}{2}

Convert element to fraction:

arccos (- \frac{3}{4}) = \frac{arccos ( - \frac{3}{4} ) 2}{2}

= \frac{arccos ( - \frac{3}{4} ) \cdot 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{arccos ( - \frac{3}{4} ) \cdot 2 - π}{2}

x \geq \frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn

\frac{π}{2} + x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn : x \leq \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn

\frac{π}{2} + x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn

Move

\frac{π}{2}

to the right side

\frac{π}{2} + x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn

Subtract

\frac{π}{2}

from both sides

\frac{π}{2} + x - \frac{π}{2} \leq 2 π - arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

Simplify

x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

x \leq 2 π - arccos (- \frac{3}{4}) + 2 πn - \frac{π}{2}

Simplify

2 π - arccos (- \frac{3}{4}) - \frac{π}{2} : \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2}

2 π - arccos (- \frac{3}{4}) - \frac{π}{2}

Convert element to fraction:

2 π = \frac{2 π 2}{2}, arccos (- \frac{3}{4}) = \frac{arccos ( - \frac{3}{4} ) 2}{2}

= \frac{2 π 2}{2} - \frac{arccos ( - \frac{3}{4} ) \cdot 2}{2} - \frac{π}{2}

Since the denominators are equal, combine the fractions:

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

= \frac{2 π 2 - arccos ( - \frac{3}{4} ) \cdot 2 - π}{2}

2 π 2 - arccos (- \frac{3}{4}) \cdot 2 - π = 3 π - 2 arccos (- \frac{3}{4})

2 π 2 - arccos (- \frac{3}{4}) \cdot 2 - π

Multiply the numbers:

2 \cdot 2 = 4

= 4 π - 2 arccos (- \frac{3}{4}) - π

Group like terms

= 4 π - π - 2 arccos (- \frac{3}{4})

Add similar elements:

4 π - π = 3 π

= 3 π - 2 arccos (- \frac{3}{4})

= \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2}

x \leq \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn

Combine the intervals

x \geq \frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn and x \leq \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn

Merge Overlapping Intervals

\frac{2 arccos ( - \frac{3}{4} ) - π}{2} + 2 πn \leq x \leq \frac{3 π - 2 arccos ( - \frac{3}{4} )}{2} + 2 πn

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