Question

WILL GIVE BRAINLIEST

Answer 1

$\bf \stackrel{\textit{testing for the x-axis symmetry, }\theta =-\theta }{r=8cos(3\theta )\implies r=8cos[3(-\theta)]}\implies r=8cos(-3\theta) \\\\\\ r=8cos(3\theta)~~\boxed{\checkmark}~~\impliedby \stackrel{\textit{trigonometry symmetry identities}}{cos(\theta)=cos(-\theta)}\\\\ -------------------------------$

$\bf \stackrel{\textit{testing for the y-axis symmetry, }\theta =\pi -\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi -\theta)]}\implies r=8cos(3\pi -3\theta)\boxed{\otimes}\\\\ -------------------------------\\\\ \stackrel{\textit{testing for the origin symmetry, }\theta =\pi +\theta }{r=8cos(3\theta )\implies r=8cos[3(\pi +\theta)]}\implies r=8cos(3\pi +3\theta)\boxed{\otimes}$

so as you can see, since the x-axis test yielded the same original expression, it has symmetry with the x-axis, or namely the "polar axis".

Answer 2

Answer:

The graph is symmetric about the x-axis.

Step-by-step explanation:

The given polar equation is

$r=8\cos 3\theta$

If the polar equation satisfy by the point (r,-θ), then it is symmetric about the x-axis.

If the polar equation satisfy by the point (-r,-θ), then it is symmetric about the y-axis.

If the polar equation satisfy by the point (-r,θ), then it is symmetric about the origin.

Substitute (r,-θ) for  (r,θ) in the given equation.

$r=8\cos 3(-\theta)$

$r=8\cos 3\theta$

$r=r$

Therefore the graph is symmetric about the x-axis.

Substitute (-r,-θ) for  (r,θ) in the given equation.

$-r=8\cos 3(-\theta)=8\cos 3\theta=r\neq -r$

The graph is not symmetric about y-axis.

Substitute (-r,θ) for  (r,θ) in the given equation.

$-r=8\cos 3\theta=r\neq -r$

The graph is not symmetric about the origin.