What is the general solution for sin(y)=0,xcos(y)=0 ?

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

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sin(y)=0,xcos(y)=0

sin (y) = 0, x cos (y) = 0

No Solution for y \in R

Solution steps

sin (y) = 0, x cos (y) = 0

General solutions for

sin (y) = 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}

y = 0 + 2 πn, y = π + 2 πn

y = 0 + 2 πn, y = π + 2 πn

Solve

y = 0 + 2 πn : y = 2 πn

y = 0 + 2 πn

0 + 2 πn = 2 πn

y = 2 πn

y = 2 πn, y = π + 2 πn

Solutions for the range

x cos (y) = 0

No Solution for y \in R

sin (6 0^{\circ} - x) - sin (6 0^{\circ} + x) = \frac{3}{2}

tan (\frac{x}{4}) + 3 = 0, 0 \leq x \leq 2 π

cot^{(2)} (x) - csc (x) - 1 = 0

4 sin (x) + 1 = 3 csc (x)

cos^{2} (x) - 3 sin (x) = 0

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