Another approach could be as follows:

let ABC be a triangle inscribed in the circle with center O and radius r.

If the area of this triangle is maximum, then vertex C should be at a maximum distance from the base AB i.e. CD must be perpendicular to AB.

Hence, ABC is an isosceles triangle.
If  BCD = , where D is the mid-point of BC, then BOD = 2

so, AB = 2 BD

= 2r sin 2

CD = CO + OD = r + r cos 2

If S be the area of the triangle ABC, then

S = (1/2) AB x CD

= (1/2) x 2r sin 2 (r + r cos 2)

ds/d = r²[sin2 (-2 sin2) + (1 + cos2)(2 cos2)]

= 2r²[cos²2 - sin²2 + cos2] = 2r²(cos4 + cos2)

For maximum and mimimum

ds/d = 0

or, cos4 + cos2 = 0

or, 2 cos3 cos = 0

so, Either cos3 = 0 or, cos =0

If cos = 0 or 3 = /2 or = /6

(d²s/d²) = -ve, for = /6

ACB = 2 = 2(/6) = /3 = ABC = BAC

so ABC is an equilateral triangle.