\( \sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3} \)

1.	\( \sin ^{-1} x+\sin ^{-1} 2 x=\frac{\pi}{3} \)
Answer» sin^-1 x + sin^-12x = \(\frac{π}{3}\) sin-1 ( x\((x { \sqrt{1-4x^2} \ } + 2x { \sqrt{1-x^2} \ }) =\frac{π}{3}\) ( ∵ sin^-1x + sin^-1y = sin^-1\((2x { \ \sqrt{1-y2} \ }+{ 4 \sqrt{1-x^2} \ }))\)^\(\) ⇒ ^{\(x { \sqrt{1-4x^2} \ } +2x { \sqrt{1-x^2} \ } =sin\frac{π}{3}=\frac{√3}{2}\)} ⇒ x² (1- 4x²) + 4 x² (1-x²) + 4x² \( { \ \sqrt{(1-4x^2)(1-x^2)} \ }=\frac{3}{4}\) (by squaring on both sides) ⇒ 16x²\( { \ \sqrt{(1-x^2)(1-4x^2)} }=3-(x^2 -4x^4+4x^2-4x^4)\) ⇒ \(16x^2 = { \sqrt{(1-x^2)(1-4x^2)} \ } =3-5x^2+8x^4\) ⇒ 256x⁴(1 - x²)(1- 4x²)= 9 + 25x⁴+ 64x⁸ - 30 x² + 48 x⁴- 80x⁶ ⇒ 256x⁴(1 - 5x²+4x⁴) = 64x⁸ - 80x⁶ + 73x⁴ - 30 x² + 9 ⇒ 256x⁴- 1280x⁶ + 1024 x⁸ = 64 x⁸ - 80 x⁶ + 73x⁴ - 30x² +9 ⇒ 960x⁸ - 1200x⁶+ 183x⁴ +30x² - 9 = 0

Answer»

sin^-1 x + sin^-12x = \(\frac{π}{3}\)

sin-1 ( x\((x { \sqrt{1-4x^2} \ } + 2x { \sqrt{1-x^2} \ }) =\frac{π}{3}\)

( ∵ sin^-1x + sin^-1y = sin^-1\((2x { \ \sqrt{1-y2} \ }+{ 4 \sqrt{1-x^2} \ }))\)^\(\)

⇒ ^{\(x { \sqrt{1-4x^2} \ } +2x { \sqrt{1-x^2} \ } =sin\frac{π}{3}=\frac{√3}{2}\)}

⇒ x² (1- 4x²) + 4 x² (1-x²) + 4x² \( { \ \sqrt{(1-4x^2)(1-x^2)} \ }=\frac{3}{4}\)

(by squaring on both sides)

⇒ 16x²\( { \ \sqrt{(1-x^2)(1-4x^2)} }=3-(x^2 -4x^4+4x^2-4x^4)\)

⇒ \(16x^2 = { \sqrt{(1-x^2)(1-4x^2)} \ } =3-5x^2+8x^4\)

⇒ 256x⁴(1 - x²)(1- 4x²)= 9 + 25x⁴+ 64x⁸ - 30 x² + 48 x⁴- 80x⁶

⇒ 256x⁴(1 - 5x²+4x⁴) = 64x⁸ - 80x⁶ + 73x⁴ - 30 x² + 9

⇒ 256x⁴- 1280x⁶ + 1024 x⁸ = 64 x⁸ - 80 x⁶ + 73x⁴ - 30x² +9

⇒ 960x⁸ - 1200x⁶+ 183x⁴ +30x² - 9 = 0

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