The circumradius of an isosceles triangle ABC is four times as that of inradius and A=B condition

Question

The circumradius of an isosceles triangle ABC is four times as that of inradius of the triangle, if A = B. Then

(1) $8 cos^2A – 8cosA + 1 = 0$

(2) $4 cos^2A – 10cosA + 1 = 0$

(3) $cos^2A – cosA – 3 = 0$

(4) $cos^2A – cosA – 8 = 0$

I am trying to use the following approach $R=\frac{abc}{4A}$, where R is Circum-radius and r is inradius and A is the area of inscribed triangle $A=rs$ ahre r=inradius and s= semiperimeter

a=a, b=a{ISOSCELES TRIANGLE} & c=c

$A=\frac{c}{2} \sqrt{(a^2-\frac{c^2}{4})}$

$s=\frac{2a+c}{2}$

$\frac{R}{r}=4$

$abcs=16A^2$

$R=\frac{abc}{4A}=\frac{a^2c}{4A}$

$16A^2=c^2(4a^2-c^2)$

$abc(\frac{2a+c}{2})=c^2(4a^2-c^2)$

$a^2(\frac{2a+c}{2})=c(4a^2-c^2)$

$2a^3+a^2c=8a^2c-2c^3$

$7a^2c-2a^3-2c^3=0$

2A+C=180

C=180-2A

sinC=sin2A

$\frac{sinA}{a}=\frac{sinC}{c}$

$c=\frac{asinC}{sinA}=2acosA$

$7a^2*2acosA-2a^3-2(2acosA)^3=0$

$7cosA-1-8cos^3A=0$

Still not able to get the answer , I presume that I am making a mistake

Answer 1

Hint:

Using this $$r=R(\cos A+\cos B+\cos C-1)$$

We have

$A=B\implies \cos B=\cos A,\cos C=\cos(\pi-A-A)=-\cos2A$

and $R=4r$

$\implies r=(4r)(\cos A+\cos A-\cos2A-1)$

Hope you can take it home from here?