The number of solutions of the equation sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x² for x ∈ [-1, 1], and

Q: Why do we split the domain at $t = 2/3$?

The Greatest Integer Function $[z]$ changes value only at integer points. By setting the arguments $t + \frac{1}{3}$ and $t - \frac{2}{3}$ to integers, we find the 'jump' points. At $t = 2/3$, both arguments hit an integer simultaneously, making it the perfect boundary for our case analysis.

Q: Wait, is $\cos^{-1}(-1)$ really $\pi$?

Yes, absolutely. Remember the unit circle? $\cos(\theta) = -1$ occurs at $\theta = \pi$ radians (or 180 degrees). Since the range of $\cos^{-1}(x)$ is $[0, \pi]$, $\pi$ is the correct and only value.

Q: I got 0 as the answer, but I feel like I missed something. How can the answer be 0?

It is a common psychological trap to feel that a math problem *must* have a solution. But in mathematics, 'no solution' is a perfectly valid and elegant answer. Trust your derivation: if the LHS is a constant $\pi$ and the RHS is $t \in [0, 1]$, they can never intersect.

Question

The number of solutions of the equation $\sin^{-1} \left[ x^2 + \frac{1}{3} \right] + \cos^{-1} \left[ x^2 - \frac{2}{3} \right] = x^2$ for $x \in [-1, 1]$, and $[x]$ denotes the greatest integer less than or equal to $x$, is :

Options:

(A) 2

(B) 0

(C) 4

(D) Infinite

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are tasked with solving the equation:

$$\sin^{-1} [x^2 + \frac{1}{3}] + \cos^{-1} [x^2 - \frac{2}{3}] = x^2$$

for the domain $x \in [-1, 1]$. Given this constraint, we observe that $x^2$ is trapped in the interval $[0, 1]$.

Let us define a new variable $t = x^2$. Our equation now transforms into:

$$\sin^{-1} [t + \frac{1}{3}] + \cos^{-1} [t - \frac{2}{3}] = t$$

### The Role of the Greatest Integer Function

The Greatest Integer Function (GIF) is a staircase function that remains constant between integers. To solve this, we must identify the points where the arguments inside the brackets become integers.

For the first term, $t + \frac{1}{3} = 1$ implies $t = \frac{2}{3}$. For the second term, $t - \frac{2}{3} = 0$ implies $t = \frac{2}{3}$.

This critical transition point, $t = \frac{2}{3}$, divides our domain $[0, 1]$ into two distinct regions: $[0, \frac{2}{3})$ and $[\frac{2}{3}, 1]$.

### Case 1: The Interval $[0, \frac{2}{3})$

In this interval, $t + \frac{1}{3}$ ranges from $\frac{1}{3}$ to strictly less than $1$. Therefore, $[t + \frac{1}{3}] = 0$.

For the second term, $t - \frac{2}{3}$ ranges from $-\frac{2}{3}$ to strictly less than $0$. Therefore, $[t - \frac{2}{3}] = -1$.

Substituting these into our equation:

$$\sin^{-1}(0) + \cos^{-1}(-1) = t$$

Since $\sin^{-1}(0) = 0$ and $\cos^{-1}(-1) = \pi$, we get $\pi = t$. However, this contradicts our assumption that $t < \frac{2}{3}$. Thus, there are no solutions in this interval.

### Case 2: The Interval $[\frac{2}{3}, 1]$

In this interval, $t + \frac{1}{3}$ ranges from $1$ to $\frac{4}{3}$. Therefore, $[t + \frac{1}{3}] = 1$.

For the second term, $t - \frac{2}{3}$ ranges from $0$ to $\frac{1}{3}$. Therefore, $[t - \frac{2}{3}] = 0$.

Substituting these into our equation:

$$\sin^{-1}(1) + \cos^{-1}(0) = t$$

Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\cos^{-1}(0) = \frac{\pi}{2}$, we get $\pi = t$. Again, this contradicts our assumption that $t \le 1$.

### Conclusion

We have exhausted all possible intervals within the domain. In both cases, the left-hand side of the equation evaluates to the constant $\pi$, while the right-hand side $t$ is restricted to the interval $[0, 1]$.

Since $\pi \approx 3.14$ can never equal a value in $[0, 1]$, there are **no solutions** to this equation.

Animated Solution for Mathematics - Inverse Trigonometric Functions: The number of solutions of the equation $sin^{- 1} [x^{2} + \frac{1}{3}] + cos^{- 1} [x^{2} - \frac{2}{3}] = x^{2}$ for $x \in [- 1, 1]$ , and $[x]$ denotes the greatest integer less than or equal to $x$ , is :

Select Answer:

Visualized Solution

Analyzing the Domain $x \in [- 1, 1]$

Critical Points for GIF Transitions

Case 1: $0 \leq t < \frac{2}{3}$

Evaluating $[t + \frac{1}{3}]$ in Case 1

Evaluating $[t - \frac{2}{3}]$ in Case 1

Finding LHS for Case 1

Checking for Solutions in Case 1

Case 2: $\frac{2}{3} \leq t \leq 1$

Evaluating $[t + \frac{1}{3}]$ in Case 2

Evaluating $[t - \frac{2}{3}]$ in Case 2

Finding LHS for Case 2

Checking for Solutions in Case 2

Final Tally of Solutions

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Role of the Greatest Integer Function

Case 1

Case 2

Conclusion

Similar Questions

The number of real solutions of $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = π /2$ is

If $k = tan (\frac{π}{4} + \frac{1}{2} cos^{- 1} (\frac{2}{3})) + tan (\frac{1}{2} sin^{- 1} (\frac{2}{3}))$ , then the number of solutions of the equation $sin^{- 1} (k x - 1) = sin^{- 1} x - cos^{- 1} x$ is ......... .

The number of real roots of the equation $tan^{- 1} x (x + 1) + sin^{- 1} x^{2} + x + 1 = \frac{π}{4}$ is :

The number of solutions of $tan^{- 1} 4 x + tan^{- 1} 6 x = \frac{π}{6}$ , where $- \frac{1}{2 6} < x < \frac{1}{2 6}$ , is equal to

Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy $sin^{- 1} (\frac{3 x}{5}) + sin^{- 1} (\frac{4 x}{5}) = sin^{- 1} x$ is equal to:

Let $S$ be the set of all solutions of the equation $cos^{- 1} (2 x) - 2 cos^{- 1} (1 - x^{2}) = π, x \in [- \frac{1}{2}, \frac{1}{2}]$ . Then $\sum_{x \in S} 2 sin^{- 1} (x^{2} - 1)$ is equal to

Considering the principal values of the inverse trigonometric functions, the sum of all the solutions of the equation $cos^{- 1} (x) - 2 sin^{- 1} (x) = cos^{- 1} (2 x)$ is equal to :

Considering only the principal values of inverse trigonometric functions, the number of positive real values of $x$ satisfying $tan^{- 1} (x) + tan^{- 1} (2 x) = \frac{π}{4}$ is :

If $sin^{- 1} (\frac{x}{5}) + cosec^{- 1} (\frac{5}{4}) = \frac{π}{2}$ , then the values of $x$ is

The value of $x$ for which $sin (cot^{- 1} (1 + x)) = cos (tan^{- 1} x)$ is

Select Answer:

Visualized Solution

Analyzing the Domain .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x∈[−1,1]

Critical Points for GIF Transitions

Case 1: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 0≤t<32​

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [t+31​] in Case 1

Evaluating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [t−32​] in Case 1