Let θ=21sin−132, then 21cos−131=(4π−θ)
k=tanθ+cotθ=sinθcosθ1=sin2θ2
k=322=3
sin−1(3x−1)=sin−1x−cos−1x
sin−1(3x−1)=2π−2cos−1x
3x−1=sin(2π−2cos−1x)
3x−1=2x2−1⇒x=0,23 (rejected)
No. of solution =1
If k=tan(4π+21cos−1(32))+tan(21sin−1(32)), then the number of solutions of the equation sin−1(kx−1)=sin−1x−cos−1x is ____
Held on 28 Jan 2026 · Verified 6 Jul 2026.
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