The number of solutions of
sinsin−1log1/2x+2cossin−1x2−32=0, is
1
2
3
none of these
The two terms on the LHS of the given equation
are meaningful, if
−1≤log1/2x≤1 and −1≤x2−32≤1⇒ 2≥x≥12 and 1≤x≤5⇒1≤x≤2
Now,
1≤x≤2⇒12≤x2≤1⇒−12≤x2−1≤0⇒−π6≤sin−1x2−1≤0⇒32≤cossin−1x2−1≤1⇒3≤2cossin−1x2−1≤2⇒3≤2cossin−1x2−1≤2
Again,
1≤x≤2⇒−1≤log1/2x≤0⇒−π2≤sin−1log1/2x≤0⇒−1≤sinsin−1log1/2x≤0⇒3−1≤sinsin−1log1/2x+cossin−1x2−32≤2⇒ sinsin−1log1/2x+2cossin−1x2−32
≠0 for any x
Hence, the given equation has no solution.