What is the general solution for 0=5*sin(2pi*(-0.25-x)) ?

The general solution for 0=5*sin(2pi*(-0.25-x)) is x=-n-0.25,x=-n-0.75

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0=5sin(2pi(-0.25-x))

0 = 5 \cdot sin (2 π \cdot (- 0.25 - x))

x = - n - 0.25, x = - n - 0.75

Solution steps

0 = 5 sin (2 π (- 0.25 - x))

Switch sides

5 sin (2 π (- 0.25 - x)) = 0

Divide both sides by

5

sin (2 π (- 0.25 - x)) = 0

General solutions for

sin (2 π (- 0.25 - x)) = 0

2 π (- 0.25 - x) = 0 + 2 πn, 2 π (- 0.25 - x) = π + 2 πn

Solve

2 π (- 0.25 - x) = 0 + 2 πn : x = - n - 0.25

Solve

2 π (- 0.25 - x) = π + 2 πn : x = - n - 0.75

x = - n - 0.25, x = - n - 0.75

cos (a) = \frac{2}{2}

csc (x) = 1.1223

x, z = sin (5 x) cos (5 y)

cot (x) = - 0.465

sin (θ) = 22

What is the general solution for 0=5*sin(2pi*(-0.25-x)) ?
The general solution for 0=5*sin(2pi*(-0.25-x)) is x=-n-0.25,x=-n-0.75