find no. of solutions

Question

the no. of solutions of the equation sin^-1(2x) - cos^-1(x) + tan^-1(2x) = π/2 , are ?

Answer 1

sin^-12x - cos^-1x+tan^-12x \(=\frac{\pi}{2}\)

⇒ tan^-12x - tan^-1\(\frac{\sqrt{1-x^2}}{x}=\frac{\pi}{2}\) - sin^-12x(∵cos^-1x = tan^-1\(\frac{\sqrt{1-x^2}}{x}\))

⇒ tan^-1\(\left(\cfrac{2x-\frac{\sqrt{1-x^2}}{x}}{1+2x\times\frac{1-x^2}{x}}\right)\) = cos^-12x[∵sin^-1x+cos^-1x\(=\frac{\pi}{2}\) and tan^-1A - tan^-1B = tan^-1\(\left(\frac{A-B}{1+AB}\right)\)]

⇒  tan^-1\(\left(\cfrac{2x^2-\sqrt{1-x^2}}{x(1+2\sqrt{1-x^2})}\right)\) = tan^-1\(\left(\frac{\sqrt{1-4x^2}}{2x}\right)\) (∵cos^-1x = tan^-1\(\frac{\sqrt{1-x^2}}{x}\))

⇒ \(\left(\cfrac{2x^2-\sqrt{1-x^2}}{x(1+2\sqrt{1-x^2})}\right)\) \(=\frac{\sqrt{1-4x^2}}{2x}\)

⇒  \(\left(\cfrac{2x^2-\sqrt{1-x^2}}{1+2\sqrt{1-x^2}}\right)\) \(=\frac{\sqrt{1-4x^2}}{2}\)
⇒ \(\frac{4x^4-4x^2\sqrt{1-x^2}+1-x^2}{1+4\sqrt{1-x^2+4(1-x^2)}}\) \(=\frac{1-4x^2}{4}\) (By squaring both sides)

⇒ 16x⁴ - 16x²\(\sqrt{1-x^2}\) + 4 - 4x² = (1 - 4x²)(5 - 4x²+4\(\sqrt{1-x^2}\))

⇒  16x⁴ - 16x²\(\sqrt{1-x^2}\) + 4 - 4x² = 5 - 20x² - 4x²+16x⁴+4\(\sqrt{1-x^2}\) - 16x²\(\sqrt{1-x^2}\)

⇒ 20x² -1 = 4\(\sqrt{1-x^2}\)

⇒ 400x⁴ - 40x²+1 = 16(1 - x²) (By squaring both sides)

⇒ 400x⁴ - 40x²+16x²+1 - 16 = 0

⇒ 400x² - 24x² - 15 = 0

⇒ x² \(=\frac{24\pm\sqrt{576+24000}}{800}\) \(=\frac{24\pm\sqrt{24576}}{800}\)

∵ x² never be negative

Therefore, x² ≠ \(\frac{24-\sqrt{24576}}{800}\) (∵ \(\sqrt{24576}\) = 15674>24)

Thus, x² = \(=\frac{24+\sqrt{24576}}{800}\)

⇒ x \(=\pm\frac{24+\sqrt{24576}}{800}\)

Hence, the number of solution of given equation is 2.