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Popular >

0<=-cos(b)0.6428<= 180

Solution

0 \leq - cos (b) 0.6428 \leq 180

Solution

False for all b \in R

Solution steps

0 \leq - cos (b) \cdot 0.6428 \leq 180

a \leq u \leq b

then

a \leq u and u \leq b

0 \leq - cos (b) \cdot 0.6428 and - cos (b) \cdot 0.6428 \leq 180

- cos (b) \cdot 0.6428 \leq 180 :

True for all

b \in R

- cos (b) \cdot 0.6428 \leq 180

Multiply both sides by

- 1

- cos (b) \cdot 0.6428 \leq 180

Multiply both sides by -1 (reverse the inequality)

(- cos (b) \cdot 0.6428) (- 1) \geq 180 (- 1)

Simplify

0.6428 cos (b) \geq - 180

0.6428 cos (b) \geq - 180

Divide both sides by

0.6428

0.6428 cos (b) \geq - 180

Divide both sides by

0.6428

\frac{0.6428 cos ( b )}{0.6428} \geq \frac{- 180}{0.6428}

Simplify

cos (b) \geq - \frac{180}{0.6428}

cos (b) \geq - \frac{180}{0.6428}

Range of

cos (b) : - 1 \leq cos (b) \leq 1

Function range definition

Range of

b :

True for all

f (x) \in R

Function range definition

The range of polynomials with odd degree is all the real numbers

True for all f (x) \in R

Since the range of the basic

cos

function is

- 1 \leq cos (b) \leq 1

and True for all

b \in R

- 1 \leq cos (b) \leq 1

cos (b) \geq - \frac{180}{0.6428} and - 1 \leq cos (b) \leq 1 : - 1 \leq cos (b) \leq 1

Let

y = cos (b)

Combine the intervals

y \geq - \frac{180}{0.6428} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y \geq - \frac{180}{0.6428} and - 1 \leq y \leq 1

The intersection of two intervals is the set of numbers which are in both intervals

y \geq - \frac{180}{0.6428}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all b

True for all b \in R

Combine the intervals

0 \leq - cos (b) \cdot 0.6428 and True for all b \in R

Merge Overlapping Intervals

False for all b \in R and True for all b \in R

The intersection of two intervals is the set of numbers which are in both intervals

False for all

b \in R

and

True for all

b \in R

False for all b \in R

False for all b \in R

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