What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi ?

Q: What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi ?

The general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi is θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(3pi)/2

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sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi

Solution

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Solution

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

Degrees

θ = 12 0^{\circ}, θ = 24 0^{\circ}, θ = 27 0^{\circ}

Solution steps

sin (2 θ) + sin (θ) + 2 cos (θ) + 1 = 0, 0 \leq θ \leq 2 π

Rewrite using trig identities

1 + sin (2 θ) + sin (θ) + 2 cos (θ)

Use the Double Angle identity:

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

= 1 + 2 sin (θ) cos (θ) + sin (θ) + 2 cos (θ)

1 + sin (θ) + 2 cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

1 + sin (θ) + 2 cos (θ) + 2 cos (θ) sin (θ)

Factor out common term

sin (θ)

= 1 + sin (θ) (1 + 2 cos (θ)) + 2 cos (θ)

Rewrite as

= (1 + 2 cos (θ)) sin (θ) + 1 \cdot (1 + 2 cos (θ))

Factor out common term

(1 + 2 cos (θ))

= (1 + 2 cos (θ)) (sin (θ) + 1)

(1 + 2 cos (θ)) (sin (θ) + 1) = 0

Solving each part separately

1 + 2 cos (θ) = 0 or sin (θ) + 1 = 0

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

1 + 2 cos (θ) = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

1 + 2 cos (θ) = 0

Subtract

1

from both sides

1 + 2 cos (θ) - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

cos (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}

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

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

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

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

sin (θ) + 1 = 0, 0 \leq θ \leq 2 π

Move

1

to the right side

sin (θ) + 1 = 0

Subtract

1

from both sides

sin (θ) + 1 - 1 = 0 - 1

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

Solutions for the range

0 \leq θ \leq 2 π

θ = \frac{3 π}{2}

Combine all the solutions

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

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Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi ?
The general solution for sin(2θ)+sin(θ)+2cos(θ)+1=0,0<= θ<= 2pi is θ=(2pi)/3 ,θ=(4pi)/3 ,θ=(3pi)/2