Question

r = 6/(sin(theta) + cos(theta))

Answer 1

\[r=\frac{6}{\sin\theta + \cos\theta}\]
What exactly do you need help with here?

Answer 2

convert from polar form to regular form

Answer 3

Okay, so you know that
\[x=r\cos\theta\\
y=r\sin\theta\\
\tan\theta=\frac{y}{x}, \text{ right?}\]

Answer 4

yes

Answer 5

And one more conversion:
\[r=\sqrt{x^2+y^2}.\]
Okay, so in the denominator, you can rewrite the original equation as
\[\sqrt{x^2+y^2}=\frac{6}{\sin{(\tan^{-1}\frac{y}{x}})+\cos{(\tan^{-1}\frac{y}{x}})}\]
Do you think you can solve that?

Answer 6

No, I don't think I can because I have tried several times and haven't been successful

Answer 7

Okay, do you know how to rewrite the terms in the denominator?

Answer 8

nope, I haven't seen anything like that before

Answer 9

Alright then. Take a look at this picture:
If you need any help understanding it, let me know.