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8sin^2(θ)-17sin(θ)<2pi,0<= θ<2pi

Solution

8 sin^{2} (θ) - 17 sin (θ) < 2 π, 0 \leq θ < 2 π

Solution

0 \leq θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) or 2 π + arcsin (\frac{- 64 π + 289 + 17}{16}) < θ < 2 π

Interval Notation

Decimal

0 \leq θ < 3.46846 \dots or 5.95631 \dots < θ < 6.28318 \dots

Solution steps

8 sin^{2} (θ) - 17 sin (θ) < 2 π, 0 \leq θ < 2 π

Let:

u = sin (θ)

8 u^{2} - 17 u < 2 π

8 u^{2} - 17 u < 2 π : - \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

8 u^{2} - 17 u < 2 π

Rewrite in standard form

8 u^{2} - 17 u < 2 π

Subtract

2 π

from both sides

8 u^{2} - 17 u - 2 π < 2 π - 2 π

Simplify

8 u^{2} - 17 u - 2 π < 0

8 u^{2} - 17 u - 2 π < 0

Complete the square

8 u^{2} - 17 u - 2 π : 8 (u - \frac{17}{16})^{2} - 2 π - \frac{289}{32}

8 u^{2} - 17 u - 2 π

Write

8 u^{2} - 17 u - 2 π

in the form:

x^{2} + 2 a x + a^{2}

Factor out

8

8 (u^{2} - \frac{17 u}{8} - \frac{π}{4})

2 a = - \frac{17}{8} : a = - \frac{17}{16}

2 a = - \frac{17}{8}

Divide both sides by

2

2 a = - \frac{17}{8}

Divide both sides by

2

\frac{2 a}{2} = \frac{- \frac{17}{8}}{2}

Simplify

\frac{2 a}{2} = \frac{- \frac{17}{8}}{2}

Simplify

\frac{2 a}{2} : a

\frac{2 a}{2}

Divide the numbers:

\frac{2}{2} = 1

= a

Simplify

\frac{- \frac{17}{8}}{2} : - \frac{17}{16}

\frac{- \frac{17}{8}}{2}

Apply the fraction rule:

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

= - \frac{\frac{17}{8}}{2}

Apply the fraction rule:

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

\frac{\frac{17}{8}}{2} = \frac{17}{8 \cdot 2}

= - \frac{17}{8 \cdot 2}

Multiply the numbers:

8 \cdot 2 = 16

= - \frac{17}{16}

a = - \frac{17}{16}

a = - \frac{17}{16}

a = - \frac{17}{16}

Add and subtract

(- \frac{17}{16})^{2}

8 (u^{2} - \frac{17 u}{8} - \frac{π}{4} + (- \frac{17}{16})^{2} - (- \frac{17}{16})^{2})

x^{2} + 2 a x + a^{2} = (x + a)^{2}

u^{2} - \frac{17}{8} u + (- \frac{17}{16})^{2} = (u - \frac{17}{16})^{2}

8 ((u - \frac{17}{16})^{2} - \frac{π}{4} - (- \frac{17}{16})^{2})

Simplify

8 (u - \frac{17}{16})^{2} - 2 π - \frac{289}{32}

8 (u - \frac{17}{16})^{2} - 2 π - \frac{289}{32} < 0

Move

2 π

to the right side

8 (u - \frac{17}{16})^{2} - 2 π - \frac{289}{32} < 0

Add

2 π

to both sides

8 (u - \frac{17}{16})^{2} - 2 π - \frac{289}{32} + 2 π < 0 + 2 π

Simplify

8 (u - \frac{17}{16})^{2} - \frac{289}{32} < 2 π

8 (u - \frac{17}{16})^{2} - \frac{289}{32} < 2 π

Move

\frac{289}{32}

to the right side

8 (u - \frac{17}{16})^{2} - \frac{289}{32} < 2 π

Add

\frac{289}{32}

to both sides

8 (u - \frac{17}{16})^{2} - \frac{289}{32} + \frac{289}{32} < 2 π + \frac{289}{32}

Simplify

8 (u - \frac{17}{16})^{2} < 2 π + \frac{289}{32}

8 (u - \frac{17}{16})^{2} < 2 π + \frac{289}{32}

Divide both sides by

8

8 (u - \frac{17}{16})^{2} < 2 π + \frac{289}{32}

Divide both sides by

8

\frac{8 ( u - \frac{17}{16} ) ^{2}}{8} < \frac{2 π}{8} + \frac{\frac{289}{32}}{8}

Simplify

\frac{8 ( u - \frac{17}{16} ) ^{2}}{8} < \frac{2 π}{8} + \frac{\frac{289}{32}}{8}

Simplify

\frac{8 ( u - \frac{17}{16} ) ^{2}}{8} : (u - \frac{17}{16})^{2}

\frac{8 ( u - \frac{17}{16} ) ^{2}}{8}

Divide the numbers:

\frac{8}{8} = 1

= (u - \frac{17}{16})^{2}

Simplify

\frac{2 π}{8} + \frac{\frac{289}{32}}{8} : \frac{π}{4} + \frac{289}{256}

\frac{2 π}{8} + \frac{\frac{289}{32}}{8}

\frac{2 π}{8} = \frac{π}{4}

\frac{2 π}{8}

Cancel the common factor:

2

= \frac{π}{4}

\frac{\frac{289}{32}}{8} = \frac{289}{256}

\frac{\frac{289}{32}}{8}

Apply the fraction rule:

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

= \frac{289}{32 \cdot 8}

Multiply the numbers:

32 \cdot 8 = 256

= \frac{289}{256}

= \frac{π}{4} + \frac{289}{256}

(u - \frac{17}{16})^{2} < \frac{π}{4} + \frac{289}{256}

(u - \frac{17}{16})^{2} < \frac{π}{4} + \frac{289}{256}

(u - \frac{17}{16})^{2} < \frac{π}{4} + \frac{289}{256}

For

u^{n} < a

, if

n

is even then

- n a < u < n a

- \frac{π}{4} + \frac{289}{256} < u - \frac{17}{16} < \frac{π}{4} + \frac{289}{256}

a < u < b

then

a < u and u < b

- \frac{π}{4} + \frac{289}{256} < u - \frac{17}{16} and u - \frac{17}{16} < \frac{π}{4} + \frac{289}{256}

- \frac{π}{4} + \frac{289}{256} < u - \frac{17}{16} : u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

- \frac{π}{4} + \frac{289}{256} < u - \frac{17}{16}

Switch sides

u - \frac{17}{16} > - \frac{π}{4} + \frac{289}{256}

Move

\frac{17}{16}

to the right side

u - \frac{17}{16} > - \frac{π}{4} + \frac{289}{256}

Add

\frac{17}{16}

to both sides

u - \frac{17}{16} + \frac{17}{16} > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Simplify

u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

u - \frac{17}{16} < \frac{π}{4} + \frac{289}{256} : u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

u - \frac{17}{16} < \frac{π}{4} + \frac{289}{256}

Move

\frac{17}{16}

to the right side

u - \frac{17}{16} < \frac{π}{4} + \frac{289}{256}

Add

\frac{17}{16}

to both sides

u - \frac{17}{16} + \frac{17}{16} < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Simplify

u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Combine the intervals

u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16} and u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Merge Overlapping Intervals

u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16} and u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

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

u > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

and

u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < u < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Substitute back

u = sin (θ)

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < sin (θ) < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

a < u < b

then

a < u and u < b

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < sin (θ) and sin (θ) < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < sin (θ) : arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn < θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn

- \frac{π}{4} + \frac{289}{256} + \frac{17}{16} < sin (θ)

Switch sides

sin (θ) > - \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

For

sin (x) > a

, if

- 1 \leq a < 1

then

arcsin (a) + 2 πn < x < π - arcsin (a) + 2 πn

arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16}) + 2 πn < θ < π - arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16}) + 2 πn

Simplify

arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16}) : arcsin (\frac{- 64 π + 289 + 17}{16})

arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16})

\frac{π}{4} + \frac{289}{256} = \frac{64 π + 289}{16}

\frac{π}{4} + \frac{289}{256}

Join

\frac{π}{4} + \frac{289}{256} : \frac{64 π + 289}{256}

\frac{π}{4} + \frac{289}{256}

Least Common Multiplier of

4, 256 : 256

4, 256

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

256 : 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

256

256

divides by

2256 = 128 \cdot 2

= 2 \cdot 128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

4

256

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 256

= 256

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

256

For

\frac{π}{4} :

multiply the denominator and numerator by

64

\frac{π}{4} = \frac{π 64}{4 \cdot 64} = \frac{π 64}{256}

= \frac{π 64}{256} + \frac{289}{256}

Since the denominators are equal, combine the fractions:

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

= \frac{π 64 + 289}{256}

= \frac{π 64 + 289}{256}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{64 π + 289}{256}

256 = 16

256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

n a^{n} = a

1 6^{2} = 16

= 16

= \frac{64 π + 289}{16}

= arcsin (- \frac{64 π + 289}{16} + \frac{17}{16})

Join

- \frac{π 64 + 289}{16} + \frac{17}{16} : \frac{- 64 π + 289 + 17}{16}

- \frac{π 64 + 289}{16} + \frac{17}{16}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 64 + 289 + 17}{16}

= arcsin (\frac{- π 64 + 289 + 17}{16})

= arcsin (\frac{- 64 π + 289 + 17}{16})

Simplify

π - arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16}) : π - arcsin (\frac{- 64 π + 289 + 17}{16})

π - arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16})

arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16}) = arcsin (\frac{- 64 π + 289 + 17}{16})

arcsin (- \frac{π}{4} + \frac{289}{256} + \frac{17}{16})

\frac{π}{4} + \frac{289}{256} = \frac{64 π + 289}{16}

\frac{π}{4} + \frac{289}{256}

Join

\frac{π}{4} + \frac{289}{256} : \frac{64 π + 289}{256}

\frac{π}{4} + \frac{289}{256}

Least Common Multiplier of

4, 256 : 256

4, 256

Least Common Multiplier (LCM)

Prime factorization of

4 : 2 \cdot 2

4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2

Prime factorization of

256 : 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

256

256

divides by

2256 = 128 \cdot 2

= 2 \cdot 128

128

divides by

2128 = 64 \cdot 2

= 2 \cdot 2 \cdot 64

64

divides by

264 = 32 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 32

32

divides by

232 = 16 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 16

16

divides by

216 = 8 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 8

8

divides by

28 = 4 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 4

4

divides by

24 = 2 \cdot 2

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

Multiply each factor the greatest number of times it occurs in either

4

256

= 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2

Multiply the numbers:

2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 256

= 256

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

256

For

\frac{π}{4} :

multiply the denominator and numerator by

64

\frac{π}{4} = \frac{π 64}{4 \cdot 64} = \frac{π 64}{256}

= \frac{π 64}{256} + \frac{289}{256}

Since the denominators are equal, combine the fractions:

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

= \frac{π 64 + 289}{256}

= \frac{π 64 + 289}{256}

Apply radical rule:

n \frac{a}{b} = \frac{n a}{n b},

assuming

a \geq 0, b \geq 0

= \frac{64 π + 289}{256}

256 = 16

256

Factor the number:

256 = 1 6^{2}

= 1 6^{2}

Apply radical rule:

n a^{n} = a

1 6^{2} = 16

= 16

= \frac{64 π + 289}{16}

= arcsin (- \frac{64 π + 289}{16} + \frac{17}{16})

Join

- \frac{π 64 + 289}{16} + \frac{17}{16} : \frac{- 64 π + 289 + 17}{16}

- \frac{π 64 + 289}{16} + \frac{17}{16}

Since the denominators are equal, combine the fractions:

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

= \frac{- π 64 + 289 + 17}{16}

= arcsin (\frac{- π 64 + 289 + 17}{16})

= π - arcsin (\frac{- 64 π + 289 + 17}{16})

arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn < θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn

sin (θ) < \frac{π}{4} + \frac{289}{256} + \frac{17}{16} :

True for all

θ \in R

sin (θ) < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

Range of

sin (θ) : - 1 \leq sin (θ) \leq 1

Function range definition

The range of the basic

sin

function is

- 1 \leq sin (θ) \leq 1

- 1 \leq sin (θ) \leq 1

sin (θ) < \frac{π}{4} + \frac{289}{256} + \frac{17}{16} and - 1 \leq sin (θ) \leq 1 : - 1 \leq sin (θ) \leq 1

Let

y = sin (θ)

Combine the intervals

y < \frac{π}{4} + \frac{289}{256} + \frac{17}{16} and - 1 \leq y \leq 1

Merge Overlapping Intervals

y < \frac{π}{4} + \frac{289}{256} + \frac{17}{16} and - 1 \leq y \leq 1

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

y < \frac{π}{4} + \frac{289}{256} + \frac{17}{16}

and

- 1 \leq y \leq 1

- 1 \leq y \leq 1

- 1 \leq y \leq 1

True for all θ

True for all θ \in R

Combine the intervals

arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn < θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn and True for all θ \in R

Merge Overlapping Intervals

arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn < θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn

Combine the intervals

arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn < θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) + 2 πn and 0 \leq θ < 2 π

0 \leq θ < π - arcsin (\frac{- 64 π + 289 + 17}{16}) or 2 π + arcsin (\frac{- 64 π + 289 + 17}{16}) < θ < 2 π

Graph

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