The solution of ^ -1 x- ^ -1 2 x= 3 is,
Mathematics · JEE Main · NTA Exams — Trigonometry
The solution of \(\sin ^{-1} x-\sin ^{-1} 2 x= \pm \frac{\pi}{3}\) is,
- \(\pm \frac{1}{3}\)
- \(\pm \frac{1}{4}\)
- \(\pm \frac{\sqrt{3}}{2}\)
- \(\pm \frac{1}{2}\)
Answer
(D) 1 2
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