What is the general solution for solvefor x,y^1=ysin(x) ?

The general solution for solvefor x,y^1=ysin(x) is x= pi/2+2pin

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solvefor x,y^1=ysin(x)

solve for

x, y^{1} = y sin (x)

x = \frac{π}{2} + 2 πn

Degrees

x = 9 0^{\circ} + 36 0^{\circ} n

Solution steps

y^{1} = y sin (x)

Switch sides

y sin (x) = y^{1}

Apply rule

a^{1} = a

y sin (x) = y

Divide both sides by

y; y \neq = 0

y sin (x) = y

Divide both sides by

y; y \neq = 0

\frac{y sin ( x )}{y} = \frac{y}{y}; y \neq = 0

Simplify

sin (x) = 1; y \neq = 0

sin (x) = 1; y \neq = 0

General solutions for

sin (x) = 1

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 = \frac{π}{2} + 2 πn

x = \frac{π}{2} + 2 πn

- 0.3 cos (θ) = cos (2.2 + θ)

2 cos^{4} (θ) + 9 sin^{2} (θ) = 5

tan (3 x) = 2.5

sin (θ) = \frac{4}{5}, 0 < θ < \frac{π}{2}, cos (2)

25 (sec^{2} (5 x)) = 0

What is the general solution for solvefor x,y^1=ysin(x) ?
The general solution for solvefor x,y^1=ysin(x) is x= pi/2+2pin