Question

sin⁻¹x+cos⁻¹2/x =π/6,Solve it.

Answer 1

method 1 :- Given, $sin^{-1}x+cos^{-1}\frac{2}{x}=\frac{\pi}{6}$

Let cos^-1(2/x) = A
cosA = 2/x
x = 2/cosA = 2secA

we know,value of secant ≥ 1 or ≤ -1
e.g., secA ≥ 1 , or, secA ≤ -1

means, x either greater than 2 or less than -2.

but according to domain of sin^-1x , value of x should be lie between -1 to 1

both are contradictions.

hence, we can't get solution of this equation.

hence, value of x doesn't exist.

method 2 :-

sin^-1x + cos^-1(2/x) = π/6

sin^-1x + sin^-1√(1 - 4/x²) = π/6

[ using , sin^-1p + sin^-1q = sin^-1(p√(1-q^2)+q√(1-p^2)]

sin^-1[x × 2/x + √(1 - x²) × √(1 - 4/x²) ] = π/6

2 + √(1 - x²) × √(1 - 4/x²) = sinπ/6

2 + √(1 - x²) × √(1 - 4/x²) = 1/2

√(1 - x²) × √(1 - 4/x²) = 1/2 - 2 = -3/2

aww... what you are observing ? LHS is positive while RHS is negative.
so, can't possible to get value of x .