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2sin(x)+3((sin(2x))/(2sin(x)))<0

Solution

2 sin (x) + 3 (\frac{sin ( 2 x )}{2 sin ( x )}) < 0

Solution

- 0.98279 \dots + π + 2 πn < x < π + 2 πn or π + 2 πn < x < - 0.98279 \dots + 2 π + 2 πn

Interval Notation

(- 0.98279 \dots + π + 2 πn, π + 2 πn) \cup (π + 2 πn, - 0.98279 \dots + 2 π + 2 πn)

Decimal Notation

2.15879 \dots + 2 πn < x < 3.14159 \dots + 2 πn or 3.14159 \dots + 2 πn < x < 5.30039 \dots + 2 πn

Solution steps

2 sin (x) + 3 \cdot \frac{sin ( 2 x )}{2 sin ( x )} < 0

Simplify

2 sin (x) + 3 \cdot \frac{sin ( 2 x )}{2 sin ( x )} : \frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )}

2 sin (x) + 3 \cdot \frac{sin ( 2 x )}{2 sin ( x )}

Multiply

3 \cdot \frac{sin ( 2 x )}{2 sin ( x )} : \frac{3 sin ( 2 x )}{2 sin ( x )}

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

Multiply fractions:

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

= \frac{sin ( 2 x ) \cdot 3}{2 sin ( x )}

= 2 sin (x) + \frac{3 sin ( 2 x )}{2 sin ( x )}

Convert element to fraction:

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

= \frac{2 sin ( x ) \cdot 2 sin ( x )}{2 sin ( x )} + \frac{sin ( 2 x ) \cdot 3}{2 sin ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{2 sin ( x ) \cdot 2 sin ( x ) + sin ( 2 x ) \cdot 3}{2 sin ( x )}

2 sin (x) \cdot 2 sin (x) + sin (2 x) \cdot 3 = 4 sin^{2} (x) + 3 sin (2 x)

2 sin (x) \cdot 2 sin (x) + sin (2 x) \cdot 3

2 sin (x) \cdot 2 sin (x) = 4 sin^{2} (x)

2 sin (x) \cdot 2 sin (x)

Multiply the numbers:

2 \cdot 2 = 4

= 4 sin (x) sin (x)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

sin (x) sin (x) = sin^{1 + 1} (x)

= 4 sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= 4 sin^{2} (x)

= 4 sin^{2} (x) + 3 sin (2 x)

= \frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )}

\frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )} < 0

Periodicity of

2 sin (x) + 3 \frac{sin ( 2 x )}{2 sin ( x )} : 2 π

The compound periodicity of the sum of periodic functions is the least common multiplier of the periods

2 sin (x), 3 \frac{sin ( 2 x )}{2 sin ( x )}

Periodicity of

2 sin (x) : 2 π

Periodicity of

a \cdot sin (b x + c) + d = \frac{periodicity of sin ( x )}{∣ b ∣}

Periodicity of

sin (x)

2 π

= \frac{2 π}{∣ 1 ∣}

Simplify

= 2 π

Periodicity of

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

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

is composed of the following functions and periods:

sin (x)

with periodicity of

2 π

The compound periodicity is:

2 π

Combine periods:

2 π, 2 π

= 2 π

Find the zeroes and undifined points of

\frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )}

for

0 \leq x < 2 π

To find the zeroes, set the inequality to zero

\frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )} = 0

\frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )} = 0, 0 \leq x < 2 π : x = - 0.98279 \dots + π, x = - 0.98279 \dots + 2 π

\frac{4 sin ^{2} ( x ) + 3 sin ( 2 x )}{2 sin ( x )} = 0, 0 \leq x < 2 π

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

4 sin^{2} (x) + 3 sin (2 x) = 0

Rewrite using trig identities

3 sin (2 x) + 4 sin^{2} (x)

Use the Double Angle identity:

sin (2 x) = 2 sin (x) cos (x)

= 3 \cdot 2 sin (x) cos (x) + 4 sin^{2} (x)

Simplify

= 6 sin (x) cos (x) + 4 sin^{2} (x)

4 sin^{2} (x) + 6 cos (x) sin (x) = 0

Factor

4 sin^{2} (x) + 6 cos (x) sin (x) : 2 sin (x) (2 sin (x) + 3 cos (x))

4 sin^{2} (x) + 6 cos (x) sin (x)

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (x) = sin (x) sin (x)

= 4 sin (x) sin (x) + 6 sin (x) cos (x)

Rewrite

6

3 \cdot 2

Rewrite

4

2 \cdot 2

= 2 \cdot 2 sin (x) sin (x) + 3 \cdot 2 sin (x) cos (x)

Factor out common term

2 sin (x)

= 2 sin (x) (2 sin (x) + 3 cos (x))

2 sin (x) (2 sin (x) + 3 cos (x)) = 0

Solving each part separately

sin (x) = 0 or 2 sin (x) + 3 cos (x) = 0

sin (x) = 0, 0 \leq x < 2 π : x = 0, x = π

sin (x) = 0, 0 \leq x < 2 π

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

2 sin (x) + 3 cos (x) = 0, 0 \leq x < 2 π : x = - arctan (\frac{3}{2}) + π, x = - arctan (\frac{3}{2}) + 2 π

2 sin (x) + 3 cos (x) = 0, 0 \leq x < 2 π

Rewrite using trig identities

2 sin (x) + 3 cos (x) = 0

Divide both sides by

cos (x), cos (x) \neq = 0

\frac{2 sin ( x ) + 3 cos ( x )}{cos ( x )} = \frac{0}{cos ( x )}

Simplify

\frac{2 sin ( x )}{cos ( x )} + 3 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

2 tan (x) + 3 = 0

2 tan (x) + 3 = 0

Move

3

to the right side

2 tan (x) + 3 = 0

Subtract

3

from both sides

2 tan (x) + 3 - 3 = 0 - 3

Simplify

2 tan (x) = - 3

2 tan (x) = - 3

Divide both sides by

2

2 tan (x) = - 3

Divide both sides by

2

\frac{2 tan ( x )}{2} = \frac{- 3}{2}

Simplify

tan (x) = - \frac{3}{2}

tan (x) = - \frac{3}{2}

Apply trig inverse properties

tan (x) = - \frac{3}{2}

General solutions for

tan (x) = - \frac{3}{2}

tan (x) = - a \Rightarrow x = arctan (- a) + πn

x = arctan (- \frac{3}{2}) + πn

x = arctan (- \frac{3}{2}) + πn

Solutions for the range

0 \leq x < 2 π

x = - arctan (\frac{3}{2}) + π, x = - arctan (\frac{3}{2}) + 2 π

Combine all the solutions

x = 0, x = π, x = - arctan (\frac{3}{2}) + π, x = - arctan (\frac{3}{2}) + 2 π

Since the equation is undefined for:

0, π

x = - arctan (\frac{3}{2}) + π, x = - arctan (\frac{3}{2}) + 2 π

Show solutions in decimal form

x = - 0.98279 \dots + π, x = - 0.98279 \dots + 2 π

Find the undefined points:

x = 0, x = π

Find the zeros of the denominator

2 sin (x) = 0

Divide both sides by

2

2 sin (x) = 0

Divide both sides by

2

\frac{2 sin ( x )}{2} = \frac{0}{2}

Simplify

sin (x) = 0

sin (x) = 0

General solutions for

sin (x) = 0

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

0 \leq x < 2 π

x = 0, x = π

0, - 0.98279 \dots + π, π, - 0.98279 \dots + 2 π

Identify the intervals

0 < x < - 0.98279 \dots + π, - 0.98279 \dots + π < x < π, π < x < - 0.98279 \dots + 2 π, - 0.98279 \dots + 2 π < x < 2 π

Summarize in a table:

4 sin^{2} (x) + 3 sin (2 x) sin (x) \frac{4 s i n ^{2} ( x ) + 3 s i n ( 2 x )}{2 s i n ( x )} x = 0 00 Undefined 0 < x < - 0.98279 \dots + π + + + x = - 0.98279 \dots + π 0 + 0 - 0.98279 \dots + π < x < π - + - x = π 00 Undefined π < x < - 0.98279 \dots + 2 π + - - x = - 0.98279 \dots + 2 π 0 - 0 - 0.98279 \dots + 2 π < x < 2 π - - + x = 2 π 00 Undefined

Identify the intervals that satisfy the required condition:

< 0

- 0.98279 \dots + π < x < π or π < x < - 0.98279 \dots + 2 π

Apply the periodicity of

2 sin (x) + 3 \frac{sin ( 2 x )}{2 sin ( x )}

- 0.98279 \dots + π + 2 πn < x < π + 2 πn or π + 2 πn < x < - 0.98279 \dots + 2 π + 2 πn

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