MHTCET 11 May 2024 Shift 2 Paper

Concept: Solve the trigonometric equation by converting all terms to $\sin x$ and $\cos x$, then check the domain restrictions.Explanation:The equation is $\tan x + \sec x = 2\cos x$.This is defined only when $\cos x \neq 0$, so $x \neq \frac{\pi}{2}, \frac{3\pi}{2}$.Rewrite $\tan x$ and $\sec x$ in terms of sine and cosine:$$\frac{\sin x}{\cos x} + \frac{1}{\cos x} = 2\cos x$$$$\Rightarrow \sin x + 1 = 2\cos^2 x$$Using $\cos^2 x = 1 - \sin^2 x$, we get:$$\sin x + 1 = 2(1 - \sin^2 x)$$$$\Rightarrow \sin x + 1 = 2(1 - \sin x)(1 + \sin x)$$$$\Rightarrow (1 + \sin x)\left[2(1 - \sin x) - 1\right] = 0$$Since $\sin x = -1$ would give $\cos x = 0$, making $\tan x$ and $\sec x$ undefined, this factor is rejected.Thus:$$2(1 - \sin x) - 1 = 0$$$$\Rightarrow 2 - 2\sin x - 1 = 0$$$$\Rightarrow \sin x = \frac{1}{2}$$In $[0, 2\pi]$, $\sin x = \frac{1}{2}$ gives:$$x = \frac{\pi}{6}, \frac{5\pi}{6}$$Both values satisfy the domain condition $\cos x \neq 0$.Answer: There are exactly $2$ solutions, so the correct option is A. 2.