What is the general solution for r=asin(3x) ?

The general solution for r=asin(3x) is x=(arcsin(r/a))/3+(2pin)/3 ,x= pi/3+(arcsin(-r/a))/3+(2pin)/3

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r=asin(3x)

r = a sin (3 x)

x = \frac{arcsin ( \frac{r}{a} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{arcsin ( - \frac{r}{a} )}{3} + \frac{2 πn}{3}

Solution steps

r = a sin (3 x)

Switch sides

a sin (3 x) = r

Divide both sides by

a; a \neq = 0

sin (3 x) = \frac{r}{a}; a \neq = 0

Apply trig inverse properties

3 x = arcsin (\frac{r}{a}) + 2 πn, 3 x = π + arcsin (- \frac{r}{a}) + 2 πn

Solve

3 x = arcsin (\frac{r}{a}) + 2 πn : x = \frac{arcsin ( \frac{r}{a} )}{3} + \frac{2 πn}{3}

Solve

3 x = π + arcsin (- \frac{r}{a}) + 2 πn : x = \frac{π}{3} + \frac{arcsin ( - \frac{r}{a} )}{3} + \frac{2 πn}{3}

x = \frac{arcsin ( \frac{r}{a} )}{3} + \frac{2 πn}{3}, x = \frac{π}{3} + \frac{arcsin ( - \frac{r}{a} )}{3} + \frac{2 πn}{3}

sin (x) = \frac{18}{25}

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

x, cos (x) = \frac{1}{4}

cos^{2} (t) = 0

sin (x) = \frac{18}{12}

What is the general solution for r=asin(3x) ?
The general solution for r=asin(3x) is x=(arcsin(r/a))/3+(2pin)/3 ,x= pi/3+(arcsin(-r/a))/3+(2pin)/3