What is the general solution for sin(2θ)=-1/2 ,θ>= 0 ?

The general solution for sin(2θ)=-1/2 ,θ>= 0 is

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sin(2θ)=-1/2 ,θ>= 0

Solution

sin (2 θ) = - \frac{1}{2}, θ \geq 0

Solution

θ = \frac{7 π}{12}, θ = \frac{11 π}{12}, θ = \frac{19 π}{12}, θ = \frac{23 π}{12}, θ = \frac{31 π}{12}, θ = \frac{35 π}{12}, θ = \frac{43 π}{12}, θ = \frac{47 π}{12}, θ = \frac{55 π}{12}, θ = \frac{59 π}{12}

Degrees

θ = 10 5^{\circ}, θ = 16 5^{\circ}, θ = 28 5^{\circ}, θ = 34 5^{\circ}, θ = 46 5^{\circ}, θ = 52 5^{\circ}, θ = 64 5^{\circ}, θ = 70 5^{\circ}, θ = 82 5^{\circ}, θ = 88 5^{\circ}

Solution steps

sin (2 θ) = - \frac{1}{2}, θ \geq 0

General solutions for

sin (2 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}

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

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

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

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

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

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

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

Solve

2 θ = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{12} + πn

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{11 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{11 π}{6}}{2} + \frac{2 πn}{2} : \frac{11 π}{12} + πn

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

\frac{\frac{11 π}{6}}{2} = \frac{11 π}{12}

\frac{\frac{11 π}{6}}{2}

Apply the fraction rule:

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

= \frac{11 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{11 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{11 π}{12} + πn

θ = \frac{7 π}{12} + πn, θ = \frac{11 π}{12} + πn

Solutions for the range

θ \geq 0

θ = \frac{7 π}{12}, θ = \frac{11 π}{12}, θ = \frac{19 π}{12}, θ = \frac{23 π}{12}, θ = \frac{31 π}{12}, θ = \frac{35 π}{12}, θ = \frac{43 π}{12}, θ = \frac{47 π}{12}, θ = \frac{55 π}{12}, θ = \frac{59 π}{12}

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)=-1/2 ,θ>= 0 ?
The general solution for sin(2θ)=-1/2 ,θ>= 0 is