What is the general solution for r=1-sin(0) ?

The general solution for r=1-sin(0) is r=1

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r=1-sin(0)

r = 1 - sin (0)

r = 1

Solution steps

r = 1 - sin (0)

Simplify

1 - sin (0) : 1

1 - sin (0)

Use the following trivial identity:

sin (0) = 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}

= 1 - 0

Simplify

r = 1

r = 1

x, sin^{2} (y) + cos (x) y = k

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

sin (6 0^{\circ}) = \frac{x}{32}

i, sin^{2} (x) = sin (x^{2})

c, sin^{3} (x) - cos^{3} (x) = sin (x) + cos (x)

What is the general solution for r=1-sin(0) ?
The general solution for r=1-sin(0) is r=1