solve for X sin-1x + sin-12x = ∏/3 - Maths - Inverse Trigonometric Functions

Question

solve for X
sin-1x + sin-12x = ∏/3

Ankush Jain · Accepted Answer

$Given, \sin^{- 1} x + \sin^{- 1} 2 x = \frac{π}{3} \Rightarrow \sin^{- 1} x + \sin^{- 1} 2 x = \sin^{- 1} [x \sqrt{1 - (2 x)^{2}} + 2 x \sqrt{1 - x^{2}}] = \sin^{- 1} [x \sqrt{1 - 4 x^{2}} + 2 x \sqrt{1 - x^{2}}] \Rightarrow \sin^{- 1} [x \sqrt{1 - 4 x^{2}} + 2 x \sqrt{1 - x^{2}}] = \frac{π}{3} \Rightarrow x \sqrt{1 - 4 x^{2}} + 2 x \sqrt{1 - x^{2}} = \sin \frac{π}{3} \Rightarrow x \sqrt{1 - 4 x^{2}} + 2 x \sqrt{1 - x^{2}} = \frac{\sqrt{3}}{2} \Rightarrow 2 x \sqrt{1 - 4 x^{2}} + 4 x \sqrt{1 - x^{2}} = \sqrt{3} \Rightarrow 2 x \sqrt{1 - 4 x^{2}} = [\sqrt{3} - 4 x \sqrt{1 - x^{2}}]$

Squaring both sides, we get

$\Rightarrow 4 x^{2} (1 - 4 x^{2}) = 3 + 16 x^{2} (1 - x^{2}) - 8 \sqrt{3} x \sqrt{1 - x^{2}} \Rightarrow 4 x^{2} - 16 x^{4} = 3 + 16 x^{2} - 16 x^{4} - 8 \sqrt{3} x \sqrt{1 - x^{2}} \Rightarrow - 12 x^{2} - 3 = - 8 \sqrt{3} x \sqrt{1 - x^{2}} \Rightarrow 12 x^{2} + 3 = 8 \sqrt{3} x \sqrt{1 - x^{2}}$

Again squaring both sides

$(12 x^{2} + 3)^{2} = 64 \times 3 x^{2} (1 - x^{2}) \Rightarrow 144 x^{4} + 9 + 72 x^{2} = 192 x^{2} - 192 x^{2} - 192 x^{4} \Rightarrow 336 x^{4} - 120 x^{2} + 9 = 0 \Rightarrow 112 x^{4} - 40 x^{2} + 3 = 0 \Rightarrow 112 x^{4} - 28 x^{2} - 12 x^{2} + 3 = 0 \Rightarrow (4 x^{2} - 1) (28 x^{2} - 3) = 0 \Rightarrow x^{2} = \frac{1}{4} or x^{2} = \frac{3}{28} \Rightarrow x = \pm \frac{1}{2} or x = \pm \frac{1}{2} \sqrt{\frac{3}{7}} Hence, x = \frac{1}{2}, \frac{- 1}{2}, \frac{1}{2} \sqrt{\frac{3}{7}}, \frac{- 1}{2} \sqrt{\frac{3}{7}} are the roots of the given equation.$