r =3cos(theta) +sin(theta)
ok so i did
r^2= (3cos(theta) +sin(theta))^2
where do i go from here
$\displaystyle r =3cos(\theta) +sin(\theta)$
multiply each side by r giving
$\displaystyle r^2 =r(3cos(\theta) +sin(\theta))$
now as $\displaystyle r^2 = x^2+y^2$ the equation becomes
$\displaystyle x^2+y^2 =r\times 3cos(\theta) +r\times sin(\theta)$
$\displaystyle x^2+y^2 = 3x +y$
$\displaystyle x^2+y^2 - 3x - y = 0$