What is the general solution for sin(x)=0,2 ?

The general solution for sin(x)=0,2 is No Solution for x\in\mathbb{R}

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sin(x)=0,2

sin (x) = 0, 2

No Solution for x \in R

Solution steps

sin (x) = 0, 2

General solutions for

sin (x) = 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}

x = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

Solutions for the range

2

No Solution for x \in R

sin (2 x) = \frac{( 9 m - 4 )}{7}

sin (u) = 0.8746

r \cdot sin (x) = 9 - r^{2} \cdot cos^{2} (a)

0 = 2 + 2 cos (θ)

cos (4 5^{\circ} + x) - cos (4 5^{\circ} - x) = 2 cos (x)

What is the general solution for sin(x)=0,2 ?
The general solution for sin(x)=0,2 is No Solution for x\in\mathbb{R}