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

The general solution for sin(θ/2)+cos(θ)=0 is θ=(7pi)/3+4pin,θ=(11pi)/3+4pin,θ=pi+4pin

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sin(θ/2)+cos(θ)=0

Solution

sin (\frac{θ}{2}) + cos (θ) = 0

Solution

θ = \frac{7 π}{3} + 4 πn, θ = \frac{11 π}{3} + 4 πn, θ = π + 4 πn

Degrees

θ = 42 0^{\circ} + 72 0^{\circ} n, θ = 66 0^{\circ} + 72 0^{\circ} n, θ = 18 0^{\circ} + 72 0^{\circ} n

Solution steps

sin (\frac{θ}{2}) + cos (θ) = 0

Let:

u = \frac{θ}{2}

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

Rewrite using trig identities

cos (2 u) + sin (u)

Use the Double Angle identity:

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

= 1 - 2 sin^{2} (u) + sin (u)

1 + sin (u) - 2 sin^{2} (u) = 0

Solve by substitution

1 + sin (u) - 2 sin^{2} (u) = 0

Let:

sin (u) = u

1 + u - 2 u^{2} = 0

1 + u - 2 u^{2} = 0 : u = - \frac{1}{2}, u = 1

1 + u - 2 u^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 u^{2} + u + 1 = 0

Solve with the quadratic formula

- 2 u^{2} + u + 1 = 0

Quadratic Equation Formula:

For

a = - 2, b = 1, c = 1

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 2 ) \cdot 1}{2 ( - 2 )}

1^{2} - 4 (- 2) \cdot 1 = 3

1^{2} - 4 (- 2) \cdot 1

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 2) \cdot 1

Apply rule

- (- a) = a

= 1 + 4 \cdot 2 \cdot 1

Multiply the numbers:

4 \cdot 2 \cdot 1 = 8

= 1 + 8

Add the numbers:

1 + 8 = 9

= 9

Factor the number:

9 = 3^{2}

= 3^{2}

Apply radical rule:

3^{2} = 3

= 3

u_{1, 2} = \frac{- 1 \pm 3}{2 ( - 2 )}

Separate the solutions

u_{1} = \frac{- 1 + 3}{2 ( - 2 )}, u_{2} = \frac{- 1 - 3}{2 ( - 2 )}

u = \frac{- 1 + 3}{2 ( - 2 )} : - \frac{1}{2}

\frac{- 1 + 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 3}{- 2 \cdot 2}

Add/Subtract the numbers:

- 1 + 3 = 2

= \frac{2}{- 2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{- 4}

Apply the fraction rule:

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

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

u = \frac{- 1 - 3}{2 ( - 2 )} : 1

\frac{- 1 - 3}{2 ( - 2 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 3}{- 2 \cdot 2}

Subtract the numbers:

- 1 - 3 = - 4

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{- 4}{- 4}

Apply the fraction rule:

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

= \frac{4}{4}

Apply rule

\frac{a}{a} = 1

= 1

The solutions to the quadratic equation are:

u = - \frac{1}{2}, u = 1

Substitute back

u = sin (u)

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

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

sin (u) = - \frac{1}{2} : u = \frac{7 π}{6} + 2 πn, u = \frac{11 π}{6} + 2 πn

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

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

sin (u) = 1 : u = \frac{π}{2} + 2 πn

sin (u) = 1

General solutions for

sin (u) = 1

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}{2} + 2 πn

u = \frac{π}{2} + 2 πn

Combine all the solutions

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

Substitute back

u = \frac{θ}{2}

\frac{θ}{2} = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{3} + 4 πn

\frac{θ}{2} = \frac{7 π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{7 π}{6} + 2 πn

Multiply both sides by

2

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

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{7 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{7 π}{6} + 2 \cdot 2 πn : \frac{7 π}{3} + 4 πn

2 \cdot \frac{7 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{7 π}{6} = \frac{7 π}{3}

2 \cdot \frac{7 π}{6}

Multiply fractions:

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

= \frac{7 π 2}{6}

Multiply the numbers:

7 \cdot 2 = 14

= \frac{14 π}{6}

Cancel the common factor:

2

= \frac{7 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{7 π}{3} + 4 πn

θ = \frac{7 π}{3} + 4 πn

θ = \frac{7 π}{3} + 4 πn

θ = \frac{7 π}{3} + 4 πn

\frac{θ}{2} = \frac{11 π}{6} + 2 πn : θ = \frac{11 π}{3} + 4 πn

\frac{θ}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

\frac{θ}{2} = \frac{11 π}{6} + 2 πn

Multiply both sides by

2

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

Simplify

\frac{2 θ}{2} = 2 \cdot \frac{11 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{11 π}{6} + 2 \cdot 2 πn : \frac{11 π}{3} + 4 πn

2 \cdot \frac{11 π}{6} + 2 \cdot 2 πn

2 \cdot \frac{11 π}{6} = \frac{11 π}{3}

2 \cdot \frac{11 π}{6}

Multiply fractions:

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

= \frac{11 π 2}{6}

Multiply the numbers:

11 \cdot 2 = 22

= \frac{22 π}{6}

Cancel the common factor:

2

= \frac{11 π}{3}

2 \cdot 2 πn = 4 πn

2 \cdot 2 πn

Multiply the numbers:

2 \cdot 2 = 4

= 4 πn

= \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

θ = \frac{11 π}{3} + 4 πn

\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

θ = \frac{7 π}{3} + 4 πn, θ = \frac{11 π}{3} + 4 πn, θ = π + 4 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin(θ/2)+cos(θ)=0 ?
The general solution for sin(θ/2)+cos(θ)=0 is θ=(7pi)/3+4pin,θ=(11pi)/3+4pin,θ=pi+4pin