Given equation is:
cos−1(2x)−2cos−1(1−x2)=π,x∈[−21,21]
It is possible if
cos−1(2x)=π ....(1)
and 2cos−1(1−x2)=0 ....(2)
From equation (1), x=−21 which is not satisfying equation (2).
So, no such x exists.
Let S be the set of all solutions of the equation cos−1(2x)−2cos−1(1−x2)=π,x∈[−21,21]. Then x∈S∑2sin−1(x2−1) is equal to
Held on 1 Feb 2023 · Verified 6 Jul 2026.
0
3−2π
π−sin−1(43)
π−2sin−1(43)
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