Q.Show that

Rearrange the given equation:

Let $\theta=\cos^{-1}\sqrt{1-x^2}$, so $\cos\theta=\sqrt{1-x^2}$. Taking the sine of both sides:

This gives the quadratic

The argument $2x$ must satisfy $|2x|\le 1$, so the admissible root is $x=\dfrac{\sqrt3-1}{2}$ (the other root $\dfrac{-1-\sqrt3}{2}\approx-1.37$ is rejected).

Then

Concept & Intuition

We have a composition of three functions: $f(x) = \sin^{-1}(\log_2(x^2/2))$.

• The outermost function, $\sin^{-1}(u)$, is defined only when $u \in [-1,1]$.
• Inside that, $\log_2(v)$ is defined only when $v > 0$.
• The innermost expression is $v = x^2/2$, which is always non‑negative, but we must enforce $v > 0$ for the log.

So the domain is the set of $x$ such that both conditions hold simultaneously. We solve step by step.

1. Inner condition: argument of $\log_2$ must be positive

So $x$ can be any real number except $0$.

2. Outer condition: argument of $\sin^{-1}$ must be in $[-1,1]$ Let $u = \log_2\!\left(\frac{x^2}{2}\right)$. We require

Since $\log_2$ is increasing, we can exponentiate with base $2$ (preserving inequalities):

That is,

3. Solve the two inequalities

   • Left inequality: $\frac{x^2}{2} \ge \frac12 \implies x^2 \ge 1 \implies |x| \ge 1$.
   • Right inequality: $\frac{x^2}{2} \le 2 \implies x^2 \le 4 \implies |x| \le 2$.

   Combining: $1 \le |x| \le 2$.

4. Combine with $x \neq 0$

   The condition $1 \le |x| \le 2$ already excludes $0$, so the only restriction is $|x| \in [1,2]$.

   In interval notation, this means $x \in [-2,-1] \cup [1,2]$.

The key to this problem is understanding that $f(x)$ is a composition: an inverse trigonometric function applied to a rational expression. To find $A$ (domain) and $B$ (range), you must work from the inside out, respecting the constraints of $\cos^{-1}$.

The inverse cosine function $\cos^{-1}(t)$ is defined only for $t \in [-1, 1]$, and its output (range) is $[0, \pi]$. So the domain $A$ is all $x$ such that $\frac{2 - x^2}{2 + x^2}$ lies in $[-1, 1]$, and the range $B$ is the set of all possible $\cos^{-1}$ values that actually occur as $x$ varies over $A$.

Let’s work through it step by step.

1. Find the domain $A$. The inner expression is $g(x) = \frac{2 - x^2}{2 + x^2}$. Since $2 + x^2 > 0$ for all real $x$, the denominator never vanishes. We need $-1 \le g(x) \le 1$. First, note that $g(x)$ is an even function (replace $x$ with $-x$ gives the same value), so we can consider $x \ge 0$ and then include negatives symmetrically. Check the lower bound: $g(x) \ge -1$?

which is always true. So the lower bound is automatically satisfied.

Now the upper bound: $g(x) \le 1$?

which is also always true (equality at $x=0$). So $g(x)$ is always between $-1$ and $1$ for every real $x$.

Therefore, the domain $A$ is all real numbers: $A = \mathbb{R}$.

2. Find the range $B$. Since $A = \mathbb{R}$, the range $B$ is the set of all values $f(x) = \cos^{-1}(g(x))$ as $x$ runs over $\mathbb{R}$. Let’s examine what values $g(x)$ can take. As $x \to \pm\infty$,

At $x = 0$, $g(0) = \frac{2}{2} = 1$.

Since $g(x)$ is continuous, it takes all values between $-1$ and $1$ as $x$ varies from $0$ to $\infty$ (and symmetrically for negative $x$). In fact, for $x \ge 0$, $g(x)$ decreases from $1$ at $x=0$ to $-1$ as $x \to \infty$, covering every value in $(-1, 1]$ exactly once.

So the output of $\cos^{-1}$ is:

• When $g = 1$, $\cos^{-1}(1) = 0$.
• When $g = -1$, $\cos^{-1}(-1) = \pi$.
• As $g$ runs through all values in $(-1, 1)$, $\cos^{-1}(g)$ runs through all values in $(0, \pi)$. Therefore, the range $B$ is $[0, \pi]$.

3. Find $A \cap B$. $A = \mathbb{R}$ and $B = [0, \pi]$. Their intersection is simply $[0, \pi]$, since every number in $[0, \pi]$ is a real number. But note the options: (A) $[0, \sqrt{2})$, (B) $[0, \pi/2)$, (C) $[0, \pi)$, (D) $(-\pi/2, \pi/2)$. The interval $[0, \pi]$ is not exactly listed — option (C) is $[0, \pi)$, which is $[0, \pi)$ (open at $\pi$). Wait — is $\pi$ included in the range? Yes, because $g(x) = -1$ is achieved only in the limit as $x \to \pm\infty$, but not at any finite $x$. So $\cos^{-1}(-1) = \pi$ is not actually attained; the range is $[0, \pi)$, not $[0, \pi]$. Let’s verify: $g(x) = -1$ would require $2 - x^2 = -2 - x^2 \implies 2 = -2$, impossible. So $\pi$ is a supremum, not a maximum. Hence $B = [0, \pi)$. Therefore, $A \cap B = [0, \pi)$.

We are given $2 \tan^{-1} x = 3 \sin^{-1} x$ with $x \neq 0$. The equation mixes two different inverse functions, so we need a common language. A natural bridge is the identity that expresses $\tan^{-1} x$ as $\sin^{-1} \frac{x}{\sqrt{1+x^2}}$ (valid for $x \ge 0$; we'll check sign later). This lets us rewrite everything in terms of $\sin^{-1}$ alone, turning the equation into an algebraic one.

1. Rewrite $\tan^{-1} x$ in terms of $\sin^{-1}$.

   For $x > 0$, we have $\tan^{-1} x = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$.

   For $x < 0$, note that $\tan^{-1} x$ is negative and $\sin^{-1} x$ is also negative, so the same identity holds with a negative argument. We'll proceed algebraically and check consistency at the end.

   So the equation becomes:

2. Take sine of both sides. Since $\sin(\sin^{-1} t) = t$ for $t \in [-1,1]$, and both sides are in the range of $\sin^{-1}$, we apply $\sin$:

3. Use double-angle and triple-angle formulas.

   Let $A = \sin^{-1} \frac{x}{\sqrt{1+x^2}}$ and $B = \sin^{-1} x$. Then:

   • $\sin(2A) = 2 \sin A \cos A = 2 \cdot \frac{x}{\sqrt{1+x^2}} \cdot \sqrt{1 - \frac{x^2}{1+x^2}} = 2 \cdot \frac{x}{\sqrt{1+x^2}} \cdot \frac{1}{\sqrt{1+x^2}} = \frac{2x}{1+x^2}$.
   • $\sin(3B) = 3 \sin B - 4 \sin^3 B = 3x - 4x^3$.

   So the equation reduces to:

4. Solve for $x$ (with $x \neq 0$). Since $x \neq 0$, we can divide both sides by $x$:

Multiply through by $1+x^2$:

Expand:

Bring all terms to one side:

5. Solve the quadratic in $x^2$. Let $y = x^2 > 0$. Then:

Using the quadratic formula:

Since $y > 0$, we take the positive root:

6. Compute $8x^2 + 1$.

We start with

Concept & Intuition:

Inverse trig equations often become algebraic after applying a trigonometric function to both sides. But we must be careful: the ranges of $\sin^{-1}$ and $\cos^{-1}$ restrict possible $x$ values. A classic trick is to isolate one inverse function and use the identity $\sin^{-1}u + \cos^{-1}u = \frac{\pi}{2}$ to rewrite the other term, making it easier to apply sine.

Step-by-step solution:

1. Rewrite $\cos^{-1}(3x)$ using the identity For any $u$ in $[-1,1]$, we have $\cos^{-1}u = \frac{\pi}{2} - \sin^{-1}u$. So

2. Simplify the equation