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(2sin(θ)cos(θ))/((3cos^2(θ)+1))>= 16/45

Solution

\frac{2 sin ( θ ) cos ( θ )}{( 3 cos ^{2} ( θ ) + 1 )} \geq \frac{16}{45}

Solution

πn \leq θ \leq \frac{π}{2} + πn

Interval Notation

[πn, \frac{π}{2} + πn]

Decimal Notation

πn \leq θ \leq 1.57079 \dots + πn

Solution steps

\frac{2 sin ( θ ) cos ( θ )}{3 cos ^{2} ( θ ) + 1} \geq \frac{16}{45}

Use the following identity:

cos^{2} (x) + sin^{2} (x) = 1

Therefore

cos^{2} (x) = 1 - sin^{2} (x)

\frac{2 sin ( θ ) cos ( θ )}{3 ( 1 - sin ^{2} ( θ ) ) + 1} \geq \frac{16}{45}

Simplify

\frac{2 sin ( θ ) cos ( θ )}{3 ( 1 - sin ^{2} ( θ ) ) + 1} : \frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4}

\frac{2 sin ( θ ) cos ( θ )}{3 ( 1 - sin ^{2} ( θ ) ) + 1}

Expand

3 (1 - sin^{2} (θ)) + 1 : - 3 sin^{2} (θ) + 4

3 (1 - sin^{2} (θ)) + 1

Expand

3 (1 - sin^{2} (θ)) : 3 - 3 sin^{2} (θ)

3 (1 - sin^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 3, b = 1, c = sin^{2} (θ)

= 3 \cdot 1 - 3 sin^{2} (θ)

Multiply the numbers:

3 \cdot 1 = 3

= 3 - 3 sin^{2} (θ)

= 3 - 3 sin^{2} (θ) + 1

Simplify

3 - 3 sin^{2} (θ) + 1 : - 3 sin^{2} (θ) + 4

3 - 3 sin^{2} (θ) + 1

Group like terms

= - 3 sin^{2} (θ) + 3 + 1

Add the numbers:

3 + 1 = 4

= - 3 sin^{2} (θ) + 4

= - 3 sin^{2} (θ) + 4

= \frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4}

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4} \geq \frac{16}{45}

Periodicity of

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4} : π

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4}

is composed of the following functions and periods:

sin (θ)

with periodicity of

2 π

The compound periodicity is:

= π

Find the zeroes and undifined points of

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4}

for

0 \leq θ < π

To find the zeroes, set the inequality to zero

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4} = 0

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

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

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

2 sin (θ) cos (θ) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) = 0

sin (θ) = 0, 0 \leq θ < π : θ = 0

sin (θ) = 0, 0 \leq θ < π

General solutions for

sin (θ) = 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}

θ = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

Solutions for the range

0 \leq θ < π

θ = 0

cos (θ) = 0, 0 \leq θ < π : θ = \frac{π}{2}

cos (θ) = 0, 0 \leq θ < π

General solutions for

cos (θ) = 0

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ < π

θ = \frac{π}{2}

Combine all the solutions

θ = 0, θ = \frac{π}{2}

Find the undefined points:

No Solution

Find the zeros of the denominator

- 3 sin^{2} (θ) + 4 = 0

Solve by substitution

- 3 sin^{2} (θ) + 4 = 0

Let:

sin (θ) = u

- 3 u^{2} + 4 = 0

- 3 u^{2} + 4 = 0 : u = \frac{2 3}{3}, u = - \frac{2 3}{3}

- 3 u^{2} + 4 = 0

Move

4

to the right side

- 3 u^{2} + 4 = 0

Subtract

4

from both sides

- 3 u^{2} + 4 - 4 = 0 - 4

Simplify

- 3 u^{2} = - 4

- 3 u^{2} = - 4

Divide both sides by

- 3

- 3 u^{2} = - 4

Divide both sides by

- 3

\frac{- 3 u ^{2}}{- 3} = \frac{- 4}{- 3}

Simplify

u^{2} = \frac{4}{3}

u^{2} = \frac{4}{3}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{4}{3}, u = - \frac{4}{3}

\frac{4}{3} = \frac{2 3}{3}

\frac{4}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

Rationalize

\frac{2}{3} : \frac{2 3}{3}

\frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{2 3}{3}

= \frac{2 3}{3}

- \frac{4}{3} = - \frac{2 3}{3}

- \frac{4}{3}

Simplify

\frac{4}{3} : \frac{2}{3}

\frac{4}{3}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{4}{3}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{2}{3}

= - \frac{2}{3}

Rationalize

- \frac{2}{3} : - \frac{2 3}{3}

- \frac{2}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{2 3}{3 3}

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{2 3}{3}

= - \frac{2 3}{3}

u = \frac{2 3}{3}, u = - \frac{2 3}{3}

Substitute back

u = sin (θ)

sin (θ) = \frac{2 3}{3}, sin (θ) = - \frac{2 3}{3}

sin (θ) = \frac{2 3}{3}, sin (θ) = - \frac{2 3}{3}

sin (θ) = \frac{2 3}{3}, 0 \leq θ < π :

No Solution

sin (θ) = \frac{2 3}{3}, 0 \leq θ < π

- 1 \leq sin (x) \leq 1

No Solution

sin (θ) = - \frac{2 3}{3}, 0 \leq θ < π :

No Solution

sin (θ) = - \frac{2 3}{3}, 0 \leq θ < π

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

No Solution for θ \in R

0, \frac{π}{2}

Identify the intervals

0 < θ < \frac{π}{2}, \frac{π}{2} < θ < π

Summarize in a table:

sin (θ) cos (θ) - 3 sin^{2} (θ) + 4 \frac{2 s i n ( θ ) c o s ( θ )}{- 3 s i n ^{2} ( θ ) + 4} θ = 0 0 + + 0 0 < θ < \frac{π}{2} + + + + θ = \frac{π}{2} + 0 + 0 \frac{π}{2} < θ < π + - + - θ = π 0 - + 0

Identify the intervals that satisfy the required condition:

\geq 0

θ = 0 or 0 < θ < \frac{π}{2} or θ = \frac{π}{2} or θ = π

Merge Overlapping Intervals

0 \leq θ \leq \frac{π}{2} or θ = π

The union of two intervals is the set of numbers which are in either interval

θ = 0

0 < θ < \frac{π}{2}

0 \leq θ < \frac{π}{2}

The union of two intervals is the set of numbers which are in either interval

0 \leq θ < \frac{π}{2}

θ = \frac{π}{2}

0 \leq θ \leq \frac{π}{2}

The union of two intervals is the set of numbers which are in either interval

0 \leq θ \leq \frac{π}{2}

θ = π

0 \leq θ \leq \frac{π}{2} or θ = π

0 \leq θ \leq \frac{π}{2} or θ = π

Apply the periodicity of

\frac{2 sin ( θ ) cos ( θ )}{- 3 sin ^{2} ( θ ) + 4}

πn \leq θ \leq \frac{π}{2} + πn

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