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Solve Sine inverse x +sine inverse 2x = π/3
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arcsin x + arcsin 2x = π/3

arcsin 2x = π/3 - arcsin x

sin[arcsin 2x] = sin[π/3 - arcsin x] (remember the left side is like sin(a-b)

2x = sinπ/3 cos(arcsin x)-cosπ/3 sin(arc sinx)

2x = √3/2 . cos(arcsin x) - (1/2)x)

but cos(arcsin x) = √(1-x²)===>2x = √3/2 .√(1-x²) - (1/2)x)

Reduce to same denominator:

(4x) = √3 .√(1-x²) - (x)===>5x = √3 .√(1-x²)

Square both sides==> 25x²=3(1-x²)

28 x² = 3 & x² = 3/28 & x =√(3/28)



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