What is the general solution for r=a(1+sin(x)) ?

The general solution for r=a(1+sin(x)) is x=arcsin(r/a-1)+2pin,x=pi+arcsin(-r/a+1)+2pin

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r=a(1+sin(x))

r = a (1 + sin (x))

x = arcsin (\frac{r}{a} - 1) + 2 πn, x = π + arcsin (- \frac{r}{a} + 1) + 2 πn

Solution steps

r = a (1 + sin (x))

Switch sides

a (1 + sin (x)) = r

Divide both sides by

a; a \neq = 0

a (1 + sin (x)) = r

Divide both sides by

a; a \neq = 0

\frac{a ( 1 + sin ( x ) )}{a} = \frac{r}{a}; a \neq = 0

Simplify

1 + sin (x) = \frac{r}{a}; a \neq = 0

1 + sin (x) = \frac{r}{a}; a \neq = 0

Move

1

to the right side

1 + sin (x) = \frac{r}{a}; a \neq = 0

Subtract

1

from both sides

1 + sin (x) - 1 = \frac{r}{a} - 1; a \neq = 0

Simplify

sin (x) = \frac{r}{a} - 1; a \neq = 0

sin (x) = \frac{r}{a} - 1; a \neq = 0

Apply trig inverse properties

sin (x) = \frac{r}{a} - 1

General solutions for

sin (x) = \frac{r}{a} - 1

sin (x) = a \Rightarrow x = arcsin (a) + 2 πn, x = π + arcsin (a) + 2 πn

x = arcsin (\frac{r}{a} - 1) + 2 πn, x = π + arcsin (- \frac{r}{a} + 1) + 2 πn

x = arcsin (\frac{r}{a} - 1) + 2 πn, x = π + arcsin (- \frac{r}{a} + 1) + 2 πn

cos (2 x) - sin^{2} (x) - 1 = 3 sin (x)

tan (Θ) = 1.4, 18 0^{\circ} <, Θ < 27 0^{\circ}

(1 + 4 cos (θ))^{2} = (3 sin (θ))

sin (θ) - \frac{cos ( θ )}{5} = 0.6377

cos (2 θ) = 0.4, 0^{\circ} \leq θ \leq 36 0^{\circ}

What is the general solution for r=a(1+sin(x)) ?
The general solution for r=a(1+sin(x)) is x=arcsin(r/a-1)+2pin,x=pi+arcsin(-r/a+1)+2pin