If k = tan ( 4 + 1 2 cos - 1 ( 2 3 ) ) + tan ( 1 2 sin - 1 ( 2 3 ) ) , then the number of solutions of the equation sin - 1 ( k x - 1 ) = sin - 1 x - cos - 1 x is: [2026]

Question

If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]

Smart Achievers · Accepted Answer

(1)

$Let θ = \frac{1}{2} \sin^{- 1} (\frac{2}{3}), then \frac{1}{2} \cos^{- 1} (\frac{1}{3}) = (\frac{π}{4} - θ)$

$\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$

$k = \frac{2}{2 / 3} = 3$

$\sin^{- 1} (3 x - 1) = \sin^{- 1} x - \cos^{- 1} x$

$\sin^{- 1} (3 x - 1) = \frac{π}{2} - 2 \cos^{- 1} x$

$3 x - 1 = \sin (\frac{π}{2} - 2 \cos^{- 1} x)$

$3 x - 1 = 2 x^{2} - 1 \Rightarrow x = 0, \frac{3}{2} (rejected)$

$No. of solution = 1$

Solution

Q.
If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]

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Solution

Q. If k=tan(π4+12cos-1(23))+tan(12sin-1(23)), then the number of solutions of the equation sin-1(kx-1)=sin-1x-cos-1x is: [2026]

Ans. (1) Let θ=12sin-1(23), then 12cos-1(13)=(π4-θ) k=tanθ+cotθ=1sinθcosθ=2sin2θ k=22/3=3 sin-1(3x-1)=sin-1x-cos-1x sin-1(3x-1)=π2-2cos-1x 3x-1=sin(π2-2cos-1x) 3x-1=2x2-1 ⇒ x=0, 32 (rejected) No. of solution=1

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Q.
If $k = \tan ((\frac{π}{4} + \frac{1}{2} \cos^{- 1} ((\frac{2}{3})))) + \tan ((\frac{1}{2} \sin^{- 1} ((\frac{2}{3})))),$ then the number of solutions of the equation $\sin^{- 1} (k x - 1) = \sin^{- 1} x - \cos^{- 1} x$ is: [2026]