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

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

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sin(θ)sin(2θ)=0,0<= θ<= 2pi

sin (θ) sin (2 θ) = 0, 0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{2}, θ = \frac{3 π}{2}

Degrees

θ = 0^{\circ}, θ = 18 0^{\circ}, θ = 36 0^{\circ}, θ = 9 0^{\circ}, θ = 27 0^{\circ}

Solution steps

sin (θ) sin (2 θ) = 0, 0 \leq θ \leq 2 π

Solving each part separately

sin (θ) = 0 or sin (2 θ) = 0

sin (θ) = 0, 0 \leq θ \leq 2 π : θ = 0, θ = π, θ = 2 π

sin (θ) = 0, 0 \leq θ \leq 2 π

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

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = π, θ = 2 π

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

sin (2 θ) = 0, 0 \leq θ \leq 2 π

General solutions for

sin (2 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}

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

2 θ = 0 + 2 πn

0 + 2 πn = 2 πn

2 θ = 2 πn

Divide both sides by

2

2 θ = 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{2 πn}{2}

Simplify

θ = πn

θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

2 θ = π + 2 πn

Divide both sides by

2

2 θ = π + 2 πn

Divide both sides by

2

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

Simplify

θ = \frac{π}{2} + πn

θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

Solutions for the range

0 \leq θ \leq 2 π

θ = 0, θ = \frac{π}{2}, θ = π, θ = \frac{3 π}{2}, θ = 2 π

Combine all the solutions

θ = 0, θ = π, θ = 2 π, θ = \frac{π}{2}, θ = \frac{3 π}{2}

csc (4 x) - 2 = 0

cos (x) = - 0.7

tan^{2} (θ) - tan (θ) = 0

cos (8 0^{\circ})

3 tan (θ) - 1 = 0

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