JEE Advanced 2023 — Paper 1 — Question 7

Simplify: $\sqrt{1+\cos 2x} = \sqrt{2} \tan^{-1}(\tan x)$.

Using $\cos 2x = 2\cos^2 x - 1$, we get $\sqrt{1+\cos 2x} = \sqrt{2} |\cos x|$, so $|\cos x| = \tan^{-1}(\tan x)$. On each interval, $\tan^{-1}(\tan x)$ simplifies: in $(-\pi/2, \pi/2)$, it equals $x$; in $(\pi/2, 3\pi/2)$, it equals $x-\pi$; in $(-3\pi/2, -\pi/2)$,

it equals $x+\pi$. Interval I: $|\cos x| = x$. Since $\cos x > 0$, $\cos x = x$ has one root in $(0, \pi/2)$.

No root for $x < 0$. One solution. Interval II: Let $y = x-\pi$, then $-\cos x = x-\pi$ becomes $\cos y = y$, which has one root $y \in (0, \pi/2)$.

Hence $x = y+\pi \in (\pi, 3\pi/2)$ gives one solution. Interval III: Let $u = x+\pi$, then $\cos u = u$ has one root $u \in (0, \pi/2)$.

Hence $x = u-\pi \in (-\pi, -\pi/2)$ gives one solution. Thus total $3$ real solutions.