Q.The number of real solutions of the equation $\sqrt{1+\cos2x}=\sqrt2\,\cos^{-1}(\cos x)$ in $\left[\frac{\pi}{2},\pi\right]$ is
(A) $0$
(B) $1$
(C) $2$
(D) Infinite

Idea: Simplify each side using a trig identity and the inverse-cosine range, then compare sizes.

Step 1 — Left side. Since $1+\cos 2x = 2\cos^2 x$,

Step 2 — Right side. For $x \in \left[\dfrac{\pi}{2},\pi\right] \subset [0,\pi]$, the principal value gives $\cos^{-1}(\cos x) = x$, so the right side is $\sqrt{2}\,x$.

Step 3 — Solve. The equation becomes

$\sqrt{2}\,|\cos x| = \sqrt{2}\,x \quad\Longrightarrow\quad |\cos x| = x.$ …

Intuition

Both sides look complicated, but each collapses to something simple on this interval: the left side is just $\sqrt{2}$ times a cosine magnitude (bounded by $\sqrt{2}$), while the right side grows like $x$. Once we see one side is trapped below $\sqrt{2}$ and the other exceeds it, no matching is possible.

Step 1 — Simplify the left side

Use the double-angle identity $\cos 2x = 2\cos^2 x - 1$, so $1 + \cos 2x = 2\cos^2 x$. Then

The absolute value matters: on $\left[\dfrac{\pi}{2},\pi\right]$ we have $\cos x \le 0$, so $|\cos x| = -\cos x$, and the left side stays $\ge 0$ as a square root must.

Step 2 — Simplify the right side

The identity $\cos^{-1}(\cos x) = x$ holds exactly when $x$ already lies in the principal range $[0,\pi]$. Here $x \in \left[\dfrac{\pi}{2},\pi\right] \subset [0,\pi]$, so

Step 3 — Reduce the equation …

Method: Counting solutions by simplifying both sides on the given interval

To count real solutions of a trig/inverse-trig equation on an interval, simplify each side to its plainest form on that interval, then compare their achievable ranges.

Steps

Step 1: Simplify each side using identities valid on the interval.

Apply identities carefully. For instance $1+\cos 2x=2\cos^2x$, so $\sqrt{1+\cos 2x}=\sqrt2\,|\cos x|$ — keep the absolute value, since a square root is non-negative. On $[0,\pi]$, $\cos^{-1}(\cos x)=x$.

Step 2: Reduce to a single comparison.

Cancel common factors to get a clean equation, e.g. $|\cos x|=x$.

Step 3: Compare the ranges of the two sides. …

Common Mistakes

Mistake 1: Writing $\sqrt{1+\cos 2x}=\sqrt2\,\cos x$ without the absolute value.

Why it's wrong: $\sqrt{2\cos^2x}=\sqrt2\,|\cos x|$, and on $\left[\tfrac{\pi}{2},\pi\right]$, $\cos x\le 0$, so dropping the modulus gives a negative left side, impossible for a square root. Correct approach: keep $\sqrt2\,|\cos x|$.

Mistake 2: Simplifying $\cos^{-1}(\cos x)$ to something other than $x$ on this interval. …