What is the general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 ?

Q: What is the general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 ?

The general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 is θ=0,θ= pi/6 ,θ= pi/3

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sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2

Solution

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

Solution

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

Degrees

θ = 0^{\circ}, θ = 3 0^{\circ}, θ = 6 0^{\circ}

Solution steps

sin (3 θ) - sin (5 θ) = sin (θ), 0 \leq θ \leq \frac{π}{2}

Subtract

sin (θ)

from both sides

sin (3 θ) - sin (5 θ) - sin (θ) = 0

Rewrite using trig identities

sin (3 θ) - sin (5 θ) - sin (θ)

Use the Sum to Product identity:

sin (s) - sin (t) = 2 sin (\frac{s - t}{2}) cos (\frac{s + t}{2})

= - sin (θ) + 2 sin (\frac{3 θ - 5 θ}{2}) cos (\frac{3 θ + 5 θ}{2})

2 sin (\frac{3 θ - 5 θ}{2}) cos (\frac{3 θ + 5 θ}{2}) = - 2 cos (4 θ) sin (θ)

2 sin (\frac{3 θ - 5 θ}{2}) cos (\frac{3 θ + 5 θ}{2})

\frac{3 θ - 5 θ}{2} = - θ

\frac{3 θ - 5 θ}{2}

Add similar elements:

3 θ - 5 θ = - 2 θ

= \frac{- 2 θ}{2}

Apply the fraction rule:

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

= - \frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= - θ

= 2 sin (- θ) cos (\frac{3 θ + 5 θ}{2})

Use the negative angle identity:

sin (- x) = - sin (x)

= 2 cos (\frac{3 θ + 5 θ}{2}) (- sin (θ))

Remove parentheses:

(- a) = - a

= - 2 cos (\frac{3 θ + 5 θ}{2}) sin (θ)

\frac{3 θ + 5 θ}{2} = 4 θ

\frac{3 θ + 5 θ}{2}

Add similar elements:

3 θ + 5 θ = 8 θ

= \frac{8 θ}{2}

Divide the numbers:

\frac{8}{2} = 4

= 4 θ

= - 2 cos (4 θ) sin (θ)

= - sin (θ) - 2 cos (4 θ) sin (θ)

- sin (θ) - 2 cos (4 θ) sin (θ) = 0

Factor

- sin (θ) - 2 cos (4 θ) sin (θ) : - sin (θ) (2 cos (4 θ) + 1)

- sin (θ) - 2 cos (4 θ) sin (θ)

Factor out common term

- sin (θ)

= - sin (θ) (1 + 2 cos (4 θ))

- sin (θ) (2 cos (4 θ) + 1) = 0

Solving each part separately

sin (θ) = 0 or 2 cos (4 θ) + 1 = 0

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

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

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 \leq \frac{π}{2}

θ = 0

2 cos (4 θ) + 1 = 0, 0 \leq θ \leq \frac{π}{2} : θ = \frac{π}{6}, θ = \frac{π}{3}

2 cos (4 θ) + 1 = 0, 0 \leq θ \leq \frac{π}{2}

Move

1

to the right side

2 cos (4 θ) + 1 = 0

Subtract

1

from both sides

2 cos (4 θ) + 1 - 1 = 0 - 1

Simplify

2 cos (4 θ) = - 1

2 cos (4 θ) = - 1

Divide both sides by

2

2 cos (4 θ) = - 1

Divide both sides by

2

\frac{2 cos ( 4 θ )}{2} = \frac{- 1}{2}

Simplify

cos (4 θ) = - \frac{1}{2}

cos (4 θ) = - \frac{1}{2}

General solutions for

cos (4 θ) = - \frac{1}{2}

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}

4 θ = \frac{2 π}{3} + 2 πn, 4 θ = \frac{4 π}{3} + 2 πn

4 θ = \frac{2 π}{3} + 2 πn, 4 θ = \frac{4 π}{3} + 2 πn

Solve

4 θ = \frac{2 π}{3} + 2 πn : θ = \frac{π}{6} + \frac{πn}{2}

4 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

4

4 θ = \frac{2 π}{3} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{6} + \frac{πn}{2}

\frac{\frac{2 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{2 π}{12}

Cancel the common factor:

2

= \frac{π}{6}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}

Solve

4 θ = \frac{4 π}{3} + 2 πn : θ = \frac{π}{3} + \frac{πn}{2}

4 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

4

4 θ = \frac{4 π}{3} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} : θ

\frac{4 θ}{4}

Divide the numbers:

\frac{4}{4} = 1

= θ

Simplify

\frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4} : \frac{π}{3} + \frac{πn}{2}

\frac{\frac{4 π}{3}}{4} + \frac{2 πn}{4}

\frac{\frac{4 π}{3}}{4} = \frac{π}{3}

\frac{\frac{4 π}{3}}{4}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{4 π}{12}

Cancel the common factor:

4

= \frac{π}{3}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{3} + \frac{πn}{2}

θ = \frac{π}{6} + \frac{πn}{2}, θ = \frac{π}{3} + \frac{πn}{2}

Solutions for the range

0 \leq θ \leq \frac{π}{2}

θ = \frac{π}{6}, θ = \frac{π}{3}

Combine all the solutions

θ = 0, θ = \frac{π}{6}, θ = \frac{π}{3}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 ?
The general solution for sin(3θ)-sin(5θ)=sin(θ),0<= θ<= pi/2 is θ=0,θ= pi/6 ,θ= pi/3