What is the general solution for 2sin(4θ)=0 ?

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

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2sin(4θ)=0

2 sin (4 θ) = 0

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

Degrees

θ = 0^{\circ} + 9 0^{\circ} n, θ = 4 5^{\circ} + 9 0^{\circ} n

Solution steps

2 sin (4 θ) = 0

Divide both sides by

2

2 sin (4 θ) = 0

Divide both sides by

2

\frac{2 sin ( 4 θ )}{2} = \frac{0}{2}

Simplify

sin (4 θ) = 0

sin (4 θ) = 0

General solutions for

sin (4 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}

4 θ = 0 + 2 πn, 4 θ = π + 2 πn

4 θ = 0 + 2 πn, 4 θ = π + 2 πn

Solve

4 θ = 0 + 2 πn : θ = \frac{πn}{2}

4 θ = 0 + 2 πn

0 + 2 πn = 2 πn

4 θ = 2 πn

Divide both sides by

4

4 θ = 2 πn

Divide both sides by

4

\frac{4 θ}{4} = \frac{2 πn}{4}

Simplify

θ = \frac{πn}{2}

θ = \frac{πn}{2}

Solve

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

4 θ = π + 2 πn

Divide both sides by

4

4 θ = π + 2 πn

Divide both sides by

4

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

Simplify

θ = \frac{π}{4} + \frac{πn}{2}

θ = \frac{π}{4} + \frac{πn}{2}

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

θ, 4 sin^{2} (\frac{θ}{2}) + 8 cos (\frac{θ}{2}) - 7 = 0

2 sin^{2} (x) - sin (x) = 0, 0 \leq x \leq 2 π

tan (x) = \frac{2}{2}

4 cos (2 x) - 3 = 0

sin^{2} (x) + 1 = 0

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