What is the general solution for solvefor x,sin(x)cos(y)=0 ?

The general solution for solvefor x,sin(x)cos(y)=0 is x=2pin,x=pi+2pin

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

solve for

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

x = 2 πn, x = π + 2 πn

Degrees

x = 0^{\circ} + 36 0^{\circ} n, x = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (x) cos (y) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

sin (x) = 0

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

3 cos (x) - 2 sin (x) = 1

\frac{sin ( 9 5 ^{\circ} )}{27} = \frac{sin ( b )}{16}

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

sinh (x) = \frac{12}{35}

sin^{2} (x) + \frac{1}{sec ( x )} = \frac{5}{4}

What is the general solution for solvefor x,sin(x)cos(y)=0 ?
The general solution for solvefor x,sin(x)cos(y)=0 is x=2pin,x=pi+2pin