Given that the inverse trigonometric functions take principal values only. Then, the number of real

Q: Wait, why do we reject the negative root here?

Actually, in this specific problem, we do not! But it is a common fear. Always check your domain. For $\sin^{-1} x$, the domain is $[-1, 1]$. Since our solutions $x = 0, 1, -1$ all fall within this interval, they are all valid. Never assume a negative root is extraneous without checking!

Q: Why do we get extra solutions when squaring?

Squaring is a non-invertible operation. When you square both sides of an equation, you essentially introduce the possibility of solutions that satisfy the squared equation but not the original one. That is why the final verification step is not optional—it is mandatory for your survival in JEE Advanced!

Q: Is $x=0$ really a solution? It feels too simple.

In the world of JEE, simplicity is often the ultimate sophistication. When you factor out $x$ from the equation, $x=0$ emerges naturally. Never discard a solution just because it feels 'too easy.' Trust the algebra, trust the process, and always verify.

Question

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

Options:

(A) 2

(B) 1

(C) 3

(D) 0

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers. Today, we are not just solving an equation; we are peeling back the layers of a trigonometric mystery. We are faced with the equation:

$$\sin^{-1} \left( \frac{3x}{5} ight) + \sin^{-1} \left( \frac{4x}{5} ight) = \sin^{-1} x$$

Before we rush into the algebra, we must respect the Gatekeeper: the **Domain**. Inverse trigonometric functions are not defined for just any value; they have strict boundaries.

For $\sin^{-1} x$ to exist, $x$ must be trapped in the interval $[-1, 1]$. Similarly, for the terms $\frac{3x}{5}$ and $\frac{4x}{5}$, we must ensure they also reside within $[-1, 1]$.

Solving $|\frac{3x}{5}| \le 1$ gives us $|x| \le \frac{5}{3}$, and $|\frac{4x}{5}| \le 1$ gives us $|x| \le \frac{5}{4}$. When we intersect these conditions, the most restrictive one wins. Thus, our playground is strictly $x \in [-1, 1]$.

### The Algebraic Dance

Now, how do we dismantle these inverse functions? The most elegant path is to take the sine of both sides. By applying the sine function, we transform the inverse trigonometric equation into an algebraic one.

On the right side, $\sin(\sin^{-1} x)$ collapses beautifully into $x$. On the left, we face $\sin(A + B)$, where $A = \sin^{-1} (\frac{3x}{5})$ and $B = \sin^{-1} (\frac{4x}{5})$.

We invoke the classic identity: $\sin(A+B) = \sin A \cos B + \cos A \sin B$. Substituting our values, we get:

$$\frac{3x}{5} \cos\left(\sin^{-1} \frac{4x}{5}ight) + \frac{4x}{5} \cos\left(\sin^{-1} \frac{3x}{5}ight) = x$$

Using the identity $\cos(\sin^{-1} 	heta) = \sqrt{1 - 	heta^2}$, we transform the cosine terms into radicals:

$$\frac{3x}{5} \sqrt{1 - \left(\frac{4x}{5}ight)^2} + \frac{4x}{5} \sqrt{1 - \left(\frac{3x}{5}ight)^2} = x$$

### The Radical Simplification

This looks intimidating, but let us simplify. The terms inside the roots become $\sqrt{1 - \frac{16x^2}{25}}$ and $\sqrt{1 - \frac{9x^2}{25}}$. Multiplying by the denominators, we get:

$$\frac{3x \sqrt{25 - 16x^2}}{25} + \frac{4x \sqrt{25 - 9x^2}}{25} = x$$

Multiplying the entire equation by $25$ clears the clutter:

$$3x \sqrt{25 - 16x^2} + 4x \sqrt{25 - 9x^2} = 25x$$

Now, look at the structure. Every term contains an $x$. We can factor it out:

$$x\left(3 \sqrt{25 - 16x^2} + 4 \sqrt{25 - 9x^2} - 25ight) = 0$$

This immediately yields our first solution: **$x = 0$**.

### The Final Push

For $x 
eq 0$, we focus on the bracketed term: $3 \sqrt{25 - 16x^2} + 4 \sqrt{25 - 9x^2} = 25$. To solve this, we isolate one radical:

$$4 \sqrt{25 - 9x^2} = 25 - 3 \sqrt{25 - 16x^2}$$

Squaring both sides is a high-stakes move, but necessary. After careful expansion and canceling the $-144x^2$ terms, we are left with a surprisingly clean result:

$$\sqrt{25 - 16x^2} = 3$$

Squaring once more gives $25 - 16x^2 = 9$, which simplifies to $16x^2 = 16$, or $x^2 = 1$. This gives us **$x = 1$** and **$x = -1$**.

Finally, we verify. Plugging $x=1$ into the original equation gives $\sin^{-1}(0.6) + \sin^{-1}(0.8) = \sin^{-1}(1)$, which is $\approx 36.87^\circ + 53.13^\circ = 90^\circ$. It holds! The same logic applies to $x = -1$ and $x = 0$.

We have found three valid solutions: **$x \in \{-1, 0, 1\}$**.

Animated Solution for Mathematics - Inverse Trigonometric Functions: 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:

Select Answer:

Visualized Solution

Analyze the Equation and Domain

Apply Sine to Both Sides

Expand using $sin (A + B)$

Simplify the Square Roots

Clear the Denominator

Identify the First Solution: $x = 0$

Isolate One Radical Term

Square Both Sides

Simplify the Algebraic Terms

Solve for the Remaining Root

Find the Values of $x$

Final Verification and Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Algebraic Dance

The Radical Simplification

The Final Push

Similar Questions

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 :

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

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

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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 :

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

If $S = {x \in R : sin^{- 1} (\frac{x + 1}{x ^{2} + 2 x + 2}) - sin^{- 1} (\frac{x}{x ^{2} + 1}) = \frac{π}{4}}$ then $\sum_{x \in S} (sin ((x^{2} + x + 5) \frac{π}{2}) - cos ((x^{2} + x + 5) π))$ is equal to _________.

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

Select Answer:

Visualized Solution

Analyze the Equation and Domain

Apply Sine to Both Sides

Expand using .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(A+B)

Simplify the Square Roots

Clear the Denominator

Identify the First Solution: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=0

Isolate One Radical Term

Square Both Sides

Simplify the Algebraic Terms

Solve for the Remaining Root

Find the Values of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Final Verification and Conclusion

The Sigma Insight: Solving Inverse Trigonometric Equations

Analyzing the Setup

The Algebraic Dance

The Radical Simplification

The Final Push

Similar Questions

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