Solve the equation for x: arcsin x+ arcsin(1-x) = arccosx

1.	Solve the equation for x: arcsin x+ arcsin(1-x) = arccosx
Answer» arc(sinx) + arc(sin(1-x)) = arc(cosx) ⇒ sin^-1x + sin^-1(1-x) = cos^-1x ⇒ sin^-1x + sin^-1(1-x) = \(\frac{\pi}{2}\)- sin^-1x (∵ sin^-1x + cos^-1x = \(\frac{\pi}{2}\)) ⇒ 2sin^-1x = \(\frac{\pi}{2}\) - sin^-1(1-x) ⇒ sin^-1(\(2x\sqrt{1-x^2}\)) = cos^-1(1-x) (∵ 2sin^-1x = sin^-1(\(2x\sqrt{1-x^2}\)) ⇒ sin^-1(\(2x\sqrt{1-x^2}\)) = sin^-1(\(\sqrt{1-(1-x)^2}\)) (∵ cos^-1x = sin^-1\(\sqrt{1-x^2}\)) ⇒ \(2x\sqrt{1-x^2}\) = \(\sqrt{1-(1-x)^2}\) ⇒ 4x²(1-x²) = 1 - (1-x²) ⇒ 4x² - 4x⁴ = 1 - (1+x²-2x) ⇒ 4x² - 4x⁴ = -x² + 2x ⇒ 4x⁴ - 5x² + 2x = 0 ⇒ x (4x³ - 5x + 2) = 0 ⇒ x (x - \(\frac{1}{2}\)) (4x² + 2x - 4) = 0 ⇒ 2x (x - \(\frac{1}{2}\))(2x² + x - 2) = 0 ⇒ x = 0,\(\frac{1}{2}\),\(\frac{-1\pm \sqrt{1+16}}{4}\) = 0,\(\frac{1}{2}\),\(\frac{-1+\sqrt{17}}{4}\), \(\frac{-1-\sqrt{17}}{4}\) ⇒ x = 0,\(\frac{1}{2}\),\(\frac{-1+\sqrt{17}}{4}\) (∵ Domain of f(x) is x ∈ [0,\(\frac{\pi}{2}\)] ∴ x ≠ \(\frac{-1-\sqrt{17}}{4}\))

Answer»

arc(sinx) + arc(sin(1-x)) = arc(cosx)

⇒ sin^-1x + sin^-1(1-x) = cos^-1x

⇒ sin^-1x + sin^-1(1-x) = \(\frac{\pi}{2}\)- sin^-1x

(∵ sin^-1x + cos^-1x = \(\frac{\pi}{2}\))

⇒ 2sin^-1x = \(\frac{\pi}{2}\) - sin^-1(1-x)

⇒ sin^-1(\(2x\sqrt{1-x^2}\)) = cos^-1(1-x)

(∵ 2sin^-1x = sin^-1(\(2x\sqrt{1-x^2}\))

⇒ sin^-1(\(2x\sqrt{1-x^2}\)) = sin^-1(\(\sqrt{1-(1-x)^2}\))

(∵ cos^-1x = sin^-1\(\sqrt{1-x^2}\))

⇒ \(2x\sqrt{1-x^2}\) = \(\sqrt{1-(1-x)^2}\)

⇒ 4x²(1-x²) = 1 - (1-x²)

⇒ 4x² - 4x⁴ = 1 - (1+x²-2x)

⇒ 4x² - 4x⁴ = -x² + 2x

⇒ 4x⁴ - 5x² + 2x = 0

⇒ x (4x³ - 5x + 2) = 0

⇒ x (x - \(\frac{1}{2}\)) (4x² + 2x - 4) = 0

⇒ 2x (x - \(\frac{1}{2}\))(2x² + x - 2) = 0

⇒ x = 0,\(\frac{1}{2}\),\(\frac{-1\pm \sqrt{1+16}}{4}\) = 0,\(\frac{1}{2}\),\(\frac{-1+\sqrt{17}}{4}\), \(\frac{-1-\sqrt{17}}{4}\)

⇒ x = 0,\(\frac{1}{2}\),\(\frac{-1+\sqrt{17}}{4}\)

(∵ Domain of f(x) is x ∈ [0,\(\frac{\pi}{2}\)]

∴ x ≠ \(\frac{-1-\sqrt{17}}{4}\))

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