Tip:A–D to answerE for explanationV for videoS to reveal answer
Let S be the set of all solutions of the equation cos−1(2x)−2cos−1(1−x2)=π,x∈[−21,21] Then ∑x∈S2sin−1(x2−1) is equal to:
π−2sin−1(43)
Explanation
cos−1(2x)−2cos−1(1−x2)=π,x∈[2−1,21]
⇒cos−1(2x)−[cos−1(2(1−x2)−1)]=π
⇒cos−1(2x)−[cos−1(−2x2+1)]=π
⇒−cos−1[−2x2+1]=π−cos−12x
Applying cos on both sides, we get
⇒x=42±4+8=42±23=21±3
∴∑2sin−1[x2−1]=∑2sin−1[2−3]=3−2π
Explanation
cos−1(2x)−2cos−1(1−x2)=π,x∈[2−1,21]
⇒cos−1(2x)−[cos−1(2(1−x2)−1)]=π
⇒cos−1(2x)−[cos−1(−2x2+1)]=π
⇒−cos−1[−2x2+1]=π−cos−12x
Applying cos on both sides, we get
⇒x=42±4+8=42±23=21±3
∴∑2sin−1[x2−1]=∑2sin−1[2−3]=3−2π
