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

The general solution for 2sin(θ)-sin(2θ)=0 is θ=2pin,θ=pi+2pin

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2sin(θ)-sin(2θ)=0

2 sin (θ) - sin (2 θ) = 0

θ = 2 πn, θ = π + 2 πn

Degrees

θ = 0^{\circ} + 36 0^{\circ} n, θ = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

2 sin (θ) - sin (2 θ) = 0

Rewrite using trig identities

- sin (2 θ) + 2 sin (θ)

Use the Double Angle identity:

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

= - 2 sin (θ) cos (θ) + 2 sin (θ)

2 sin (θ) - 2 cos (θ) sin (θ) = 0

Factor

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

2 sin (θ) - 2 cos (θ) sin (θ)

Factor out common term

- 2 sin (θ)

= - 2 sin (θ) (- 1 + cos (θ))

- 2 sin (θ) (cos (θ) - 1) = 0

Solving each part separately

sin (θ) = 0 or cos (θ) - 1 = 0

sin (θ) = 0 : θ = 2 πn, θ = π + 2 πn

sin (θ) = 0

General solutions for

sin (θ) = 0

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 = 0 + 2 πn, θ = π + 2 πn

θ = 0 + 2 πn, θ = π + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn, θ = π + 2 πn

cos (θ) - 1 = 0 : θ = 2 πn

cos (θ) - 1 = 0

Move

1

to the right side

cos (θ) - 1 = 0

Add

1

to both sides

cos (θ) - 1 + 1 = 0 + 1

Simplify

cos (θ) = 1

cos (θ) = 1

General solutions for

cos (θ) = 1

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 = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

3 tan^{2} (x) + 4 tan (x) - 4 = 0

csc^{2} (θ) + 2 cot (θ) = 0

sin (x) cos (x) = - 10 cos (x)

tan^{2} (2 x) = 1

8 cos (x) tan (x) = tan (x)

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