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}

Solution steps

2 sin (4 θ) = 0

Divide both sides by

2

sin (4 θ) = 0

General solutions for

sin (4 θ) = 0

4 θ = 0 + 2 πn, 4 θ = π + 2 πn

Solve

4 θ = 0 + 2 πn : θ = \frac{πn}{2}

Solve

4 θ = π + 2 πn : 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