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Popular >

2cos^2(a)-1>=-1/8

Solution

2 cos^{2} (a) - 1 \geq - \frac{1}{8}

Solution

- arccos (\frac{7}{4}) + 2 πn \leq a \leq arccos (\frac{7}{4}) + 2 πn or arccos (- \frac{7}{4}) + 2 πn \leq a \leq 2 π - arccos (- \frac{7}{4}) + 2 πn

Interval Notation

[- arccos (\frac{7}{4}) + 2 πn, arccos (\frac{7}{4}) + 2 πn] \cup [arccos (- \frac{7}{4}) + 2 πn, 2 π - arccos (- \frac{7}{4}) + 2 πn]

Decimal Notation

- 0.84806 \dots + 2 πn \leq a \leq 0.84806 \dots + 2 πn or 2.29353 \dots + 2 πn \leq a \leq 3.98965 \dots + 2 πn

Solution steps

2 cos^{2} (a) - 1 \geq - \frac{1}{8}

Move

1

to the right side

2 cos^{2} (a) - 1 \geq - \frac{1}{8}

Add

1

to both sides

2 cos^{2} (a) - 1 + 1 \geq - \frac{1}{8} + 1

Simplify

2 cos^{2} (a) - 1 + 1 \geq - \frac{1}{8} + 1

Simplify

2 cos^{2} (a) - 1 + 1 : 2 cos^{2} (a)

2 cos^{2} (a) - 1 + 1

Add similar elements:

- 1 + 1 \geq 0

= 2 cos^{2} (a)

Simplify

- \frac{1}{8} + 1 : \frac{7}{8}

- \frac{1}{8} + 1

Convert element to fraction:

1 = \frac{1 \cdot 8}{8}

= \frac{1 \cdot 8}{8} - \frac{1}{8}

Since the denominators are equal, combine the fractions:

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

= \frac{1 \cdot 8 - 1}{8}

1 \cdot 8 - 1 = 7

1 \cdot 8 - 1

Multiply the numbers:

1 \cdot 8 = 8

= 8 - 1

Subtract the numbers:

8 - 1 = 7

= 7

= \frac{7}{8}

2 cos^{2} (a) \geq \frac{7}{8}

2 cos^{2} (a) \geq \frac{7}{8}

2 cos^{2} (a) \geq \frac{7}{8}

Divide both sides by

2

2 cos^{2} (a) \geq \frac{7}{8}

Divide both sides by

2

\frac{2 cos ^{2} ( a )}{2} \geq \frac{\frac{7}{8}}{2}

Simplify

\frac{2 cos ^{2} ( a )}{2} \geq \frac{\frac{7}{8}}{2}

Simplify

\frac{2 cos ^{2} ( a )}{2} : cos^{2} (a)

\frac{2 cos ^{2} ( a )}{2}

Divide the numbers:

\frac{2}{2} = 1

= cos^{2} (a)

Simplify

\frac{\frac{7}{8}}{2} : \frac{7}{16}

\frac{\frac{7}{8}}{2}

Apply the fraction rule:

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

= \frac{7}{8 \cdot 2}

Multiply the numbers:

8 \cdot 2 = 16

= \frac{7}{16}

cos^{2} (a) \geq \frac{7}{16}

cos^{2} (a) \geq \frac{7}{16}

cos^{2} (a) \geq \frac{7}{16}

For

u^{n} \geq a

, if

n

is even then

cos (a) \leq - \frac{7}{16} or cos (a) \geq \frac{7}{16}

\frac{7}{16} = \frac{7}{4}

\frac{7}{16}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{7}{16}

16 = 4

16

Factor the number:

16 = 4^{2}

= 4^{2}

Apply radical rule:

4^{2} = 4

= 4

= \frac{7}{4}

cos (a) \leq - \frac{7}{4} or cos (a) \geq \frac{7}{4}

cos (a) \leq - \frac{7}{4} : arccos (- \frac{7}{4}) + 2 πn \leq a \leq 2 π - arccos (- \frac{7}{4}) + 2 πn

cos (a) \leq - \frac{7}{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{7}{4}) + 2 πn \leq a \leq 2 π - arccos (- \frac{7}{4}) + 2 πn

cos (a) \geq \frac{7}{4} : - arccos (\frac{7}{4}) + 2 πn \leq a \leq arccos (\frac{7}{4}) + 2 πn

cos (a) \geq \frac{7}{4}

For

cos (x) \geq a

, if

- 1 < a < 1

then

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

- arccos (\frac{7}{4}) + 2 πn \leq a \leq arccos (\frac{7}{4}) + 2 πn

Combine the intervals

arccos (- \frac{7}{4}) + 2 πn \leq a \leq 2 π - arccos (- \frac{7}{4}) + 2 πn or - arccos (\frac{7}{4}) + 2 πn \leq a \leq arccos (\frac{7}{4}) + 2 πn

Merge Overlapping Intervals

- arccos (\frac{7}{4}) + 2 πn \leq a \leq arccos (\frac{7}{4}) + 2 πn or arccos (- \frac{7}{4}) + 2 πn \leq a \leq 2 π - arccos (- \frac{7}{4}) + 2 πn

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