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

The general solution for sin(θ)-cos(θ/2)=0 is θ=pi+4pin,θ=3pi+4pin,θ= pi/3+4pin,θ=(5pi)/3+4pin

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Popular Trigonometry >

sin(θ)-cos(θ/2)=0

Solution

sin (θ) - cos (\frac{θ}{2}) = 0

Solution

θ = π + 4 πn, θ = 3 π + 4 πn, θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn

Degrees

θ = 18 0^{\circ} + 72 0^{\circ} n, θ = 54 0^{\circ} + 72 0^{\circ} n, θ = 6 0^{\circ} + 72 0^{\circ} n, θ = 30 0^{\circ} + 72 0^{\circ} n

Solution steps

sin (θ) - cos (\frac{θ}{2}) = 0

Let:

u = \frac{θ}{2}

sin (2 u) - cos (u) = 0

Rewrite using trig identities

- cos (u) + sin (2 u)

Use the Double Angle identity:

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

= - cos (u) + 2 sin (u) cos (u)

- cos (u) + 2 cos (u) sin (u) = 0

Factor

- cos (u) + 2 cos (u) sin (u) : cos (u) (2 sin (u) - 1)

- cos (u) + 2 cos (u) sin (u)

Factor out common term

cos (u)

= cos (u) (- 1 + 2 sin (u))

cos (u) (2 sin (u) - 1) = 0

Solving each part separately

cos (u) = 0 or 2 sin (u) - 1 = 0

cos (u) = 0 : u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn

cos (u) = 0

General solutions for

cos (u) = 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}

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

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

2 sin (u) - 1 = 0 : u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

2 sin (u) - 1 = 0

Move

1

to the right side

2 sin (u) - 1 = 0

Add

1

to both sides

2 sin (u) - 1 + 1 = 0 + 1

Simplify

2 sin (u) = 1

2 sin (u) = 1

Divide both sides by

2

2 sin (u) = 1

Divide both sides by

2

\frac{2 sin ( u )}{2} = \frac{1}{2}

Simplify

sin (u) = \frac{1}{2}

sin (u) = \frac{1}{2}

General solutions for

sin (u) = \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}

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Combine all the solutions

u = \frac{π}{2} + 2 πn, u = \frac{3 π}{2} + 2 πn, u = \frac{π}{6} + 2 πn, u = \frac{5 π}{6} + 2 πn

Substitute back

u = \frac{θ}{2}

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

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{2} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{π}{2} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

2 \cdot \frac{π}{2} + 2 \cdot 2 πn

2 \cdot \frac{π}{2} = π

2 \cdot \frac{π}{2}

Multiply fractions:

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

= \frac{π 2}{2}

Cancel the common factor:

2

= π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

θ = π + 4 πn

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

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{3 π}{2} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

2 \cdot \frac{3 π}{2} + 2 \cdot 2 πn

2 \cdot \frac{3 π}{2} = 3 π

2 \cdot \frac{3 π}{2}

Multiply fractions:

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

= \frac{3 π 2}{2}

Cancel the common factor:

2

= 3 π

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= 3 π + 4 πn

θ = 3 π + 4 πn

θ = 3 π + 4 πn

θ = 3 π + 4 πn

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

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{π}{6} + 2 πn

Multiply both sides by

2

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

Simplify

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

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

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

2 \cdot \frac{π}{6} + 2 \cdot 2 πn

2 \cdot \frac{π}{6} = \frac{π}{3}

2 \cdot \frac{π}{6}

Multiply fractions:

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

= \frac{π 2}{6}

Cancel the common factor:

2

= \frac{π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

θ = \frac{π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{3} + 4 πn

\frac{θ}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{5 π}{6} + 2 πn

Multiply both sides by

2

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn : \frac{5 π}{3} + 4 πn

2 \cdot \frac{5 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{5 π}{6} = \frac{5 π}{3}

2 \cdot \frac{5 π}{6}

Multiply fractions:

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

= \frac{5 π 2}{6}

Multiply the numbers:

5 \cdot 2 = 10

= \frac{10 π}{6}

Cancel the common factor:

2

= \frac{5 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

θ = \frac{5 π}{3} + 4 πn

θ = π + 4 πn, θ = 3 π + 4 πn, θ = \frac{π}{3} + 4 πn, θ = \frac{5 π}{3} + 4 πn

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)-cos(θ/2)=0 ?
The general solution for sin(θ)-cos(θ/2)=0 is θ=pi+4pin,θ=3pi+4pin,θ= pi/3+4pin,θ=(5pi)/3+4pin