Hard Calc Problem...Idiot can't do it...help me?

Question

Rewrite the given equation using the substitution x=rcosA and y=rsinA. Simplify your answer.

x^2+y^2+3x=0

Answer 1

r^2(Cos A)^2 + r^2(Sin A)^2 + 3r Cos A =0,r^2(Sin^2A+Cos^2A)+3rCosA=0, then r^2 = -3rCos A,r=0,or r=-3CosA

Answer 2

(r cosA)² + (r sinA)² + 3(r cosA) = 0

r (cos²A + sin²A) + 3r cosA = 0

Use identity cos²A + sin²A = 1

r + 3r cosA = 0

r(1 + 3cosA) = 0

Answer 3

x^2+y^2 + 3x = r^2 cos^2 A + r^2 sin ^2 A + 3r cos A = 0

or r^2 + 3r cos A = 0

r(r+ 3cos A) = 0

math kp

Answer 4

Just rewrite it.

x^2 + y^2 + 3x = 0   with x = rcosA  and  y = rsin A

(rcosA)^2 + (rsinA)^2 + 3rcosA = 0

  r^2 ( (cosA)^2 + (sinA)^2 ) + 3rcosA = 0

  r^2 (1) + 3rcosA = 0

  r^2 + 3r cosA = 0

   r (r + 3cosA) = 0

So, r = 0 or r + 3cosA = 0 (in which case r = - 3 cos A)

Note:  Some may want to complete the square with the x terms first.

Answer 5

Plug them in to get

x^2= r^2 cos^2A

y^2=r^2 sin^2A

3x= 3r cosA

You can simplify by recognizing cos^2A+sin^2A=1.

You can clean this up and show your answer.

This is a calc problem?  They don't make calc the way it used to be.