<s>[INST] The circumradius of an isosceles triangle ABC is four times as that of inradius and A=B condition 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
 [/INST] 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?
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