What is the general solution for 2sin(4θ)+1=0 ?

The general solution for 2sin(4θ)+1=0 is θ=(7pi}{24}+(pin)/2 ,θ=(11pi)/(24)+\frac{pin)/2

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2sin(4θ)+1=0

Solution

2 sin (4 θ) + 1 = 0

Solution

θ = \frac{7 π}{24} + \frac{πn}{2}, θ = \frac{11 π}{24} + \frac{πn}{2}

Degrees

θ = 52. 5^{\circ} + 9 0^{\circ} n, θ = 82. 5^{\circ} + 9 0^{\circ} n

Solution steps

2 sin (4 θ) + 1 = 0

Move

1

to the right side

2 sin (4 θ) + 1 = 0

Subtract

1

from both sides

2 sin (4 θ) + 1 - 1 = 0 - 1

Simplify

2 sin (4 θ) = - 1

2 sin (4 θ) = - 1

Divide both sides by

2

2 sin (4 θ) = - 1

Divide both sides by

2

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

Simplify

sin (4 θ) = - \frac{1}{2}

sin (4 θ) = - \frac{1}{2}

General solutions for

sin (4 θ) = - \frac{1}{2}

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}

4 θ = \frac{7 π}{6} + 2 πn, 4 θ = \frac{11 π}{6} + 2 πn

4 θ = \frac{7 π}{6} + 2 πn, 4 θ = \frac{11 π}{6} + 2 πn

Solve

4 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{24} + \frac{πn}{2}

4 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

4

4 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{7 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{7 π}{6}}{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{7 π}{6}}{4} + \frac{2 πn}{4} : \frac{7 π}{24} + \frac{πn}{2}

\frac{\frac{7 π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{7 π}{6}}{4} = \frac{7 π}{24}

\frac{\frac{7 π}{6}}{4}

Apply the fraction rule:

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

= \frac{7 π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{7 π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{7 π}{24} + \frac{πn}{2}

θ = \frac{7 π}{24} + \frac{πn}{2}

θ = \frac{7 π}{24} + \frac{πn}{2}

θ = \frac{7 π}{24} + \frac{πn}{2}

Solve

4 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{24} + \frac{πn}{2}

4 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

4

4 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{\frac{11 π}{6}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 θ}{4} = \frac{\frac{11 π}{6}}{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{11 π}{6}}{4} + \frac{2 πn}{4} : \frac{11 π}{24} + \frac{πn}{2}

\frac{\frac{11 π}{6}}{4} + \frac{2 πn}{4}

\frac{\frac{11 π}{6}}{4} = \frac{11 π}{24}

\frac{\frac{11 π}{6}}{4}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 4}

Multiply the numbers:

6 \cdot 4 = 24

= \frac{11 π}{24}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{11 π}{24} + \frac{πn}{2}

θ = \frac{11 π}{24} + \frac{πn}{2}

θ = \frac{11 π}{24} + \frac{πn}{2}

θ = \frac{11 π}{24} + \frac{πn}{2}

θ = \frac{7 π}{24} + \frac{πn}{2}, θ = \frac{11 π}{24} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(4θ)+1=0 ?
The general solution for 2sin(4θ)+1=0 is θ=(7pi}{24}+(pin)/2 ,θ=(11pi)/(24)+\frac{pin)/2