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Step-by-Step Solution
Step 1: Understand the Problem
We need to find the number of real solutions (for x in the interval [−1, 1]) of the equation
$ \sin^{-1}\Big(\big[x^2+\tfrac13\big]\Big) \;+\; \cos^{-1}\Big(\big[x^2 - \tfrac23\big]\Big) \;=\; x^2 ,$
where $[\,u\,]$ denotes the greatest integer less than or equal to u.
Step 2: Analyze the Possible Integer Values of the Floor Expressions
The expression $x^2 + \tfrac13$ lies between $\tfrac13$ (when $x^2=0$) and $1+\tfrac13=\tfrac43$ (when $x^2=1$).
If $x^2+\tfrac13 < 1 \implies x^2 < \tfrac23$, the floor $\big[x^2+\tfrac13\big] = 0.$
If $x^2+\tfrac13 \ge 1 \implies x^2 \ge \tfrac23$, the floor $\big[x^2+\tfrac13\big] = 1.$
The expression $x^2 - \tfrac23$ lies between $-\tfrac23$ (when $x^2=0$) and $1-\tfrac23=\tfrac13$ (when $x^2=1$).
If $x^2 - \tfrac23 < 0 \implies x^2 < \tfrac23$, the floor $\big[x^2 - \tfrac23\big] = -1.$
If $x^2 - \tfrac23 \ge 0 \implies x^2 \ge \tfrac23$, the floor $\big[x^2 - \tfrac23\big] = 0.$
Step 3: Break Into Two Main Cases Based on $x^2$
Case A: $|x| < \sqrt{\tfrac{2}{3}}$
Here, $x^2 < \tfrac{2}{3}$. Then:
$\big[x^2 + \tfrac13\big] = 0.$
$\big[x^2 - \tfrac23\big] = -1.$
Hence, the given equation becomes:
$\sin^{-1}(0) \;+\; \cos^{-1}(-1) \;=\; x^2\,.$
We know that $\sin^{-1}(0) = 0$ and $\cos^{-1}(-1) = \pi.$ Thus,
$0 + \pi = x^2 \quad\Longrightarrow\quad x^2 = \pi.
Since $\pi \approx 3.14159$ is not possible for $x^2$ when $x \in [-1,1]$ (because $x^2 \le 1$), this case yields no solutions.
Case B: $|x| \ge \sqrt{\tfrac{2}{3}}$
Here, $x^2 \ge \tfrac{2}{3}$. Then:
$\big[x^2 + \tfrac13\big] = 1.$
$\big[x^2 - \tfrac23\big] = 0.$
The equation becomes:
$\sin^{-1}(1) \;+\; \cos^{-1}(0) \;=\; x^2\,.
We know
$\sin^{-1}(1) = \tfrac{\pi}{2}$ and $\cos^{-1}(0) = \tfrac{\pi}{2}.$
Hence,
$\tfrac{\pi}{2} + \tfrac{\pi}{2} = x^2 \quad\Longrightarrow\quad \pi = x^2.
Again, $x^2 = \pi$ is not possible for $x^2 \le 1$. Consequently, there is no solution in this case either.
Step 4: Conclusion
Since neither of the two intervals (Case A or Case B) produced a valid solution within x ∈ [−1,1],
the original equation has no real solution. Therefore, the number of solutions is 0.
