JEE Main · 2022 · 29 Jun 2022 · Shift 1 · Trigonometric Equations

Writing the equation as ${\cos }^2\theta =2\theta +\sqrt{2}$ shows the right side rises linearly from $0$ at $\theta =-\frac{\sqrt{2}}{2}$ to $1$ at $\theta =\frac{1-\sqrt{2}}{2}$, exactly the range in which ${\cos }^2\theta$ lies, so any solution must fall inside this short interval. Setting $g(\theta )=2\theta +\sqrt{2}-{\cos }^2\theta$, its derivative $g^{\prime }(\theta )=2+\sin 2\theta$ is always positive, so $g$ is strictly increasing; as it changes sign across the interval it vanishes exactly once. The equation therefore has a single real solution.