JEE Main 2024 (Online) 5th April Morning ShiftQuestion 1
Suppose $\theta\in\left[0,\dfrac{\pi}{4}\right]$ is a solution of $4\cos\theta-3\sin\theta=1$. Then $\cos\theta$ is equal to :
Trigonometric Ratio & Identities
Solution
Concept. Isolate $3\sin\theta$, square, and keep the root that fits $\theta\in\left[0,\frac\pi4\right]$. Formulas used. (1) $\sin\theta=\sqrt{1-\cos^2\theta}$ for $\theta\in\left[0,\frac\pi4\right]$Step 1 - Square. With $c=\cos\theta$, $3\sqrt{1-c^2}=4c-1$. Squaring: $$9-9c^2=16c^2-8c+1\ \Longrightarrow\ 25c^2-8c-8=0.$$ Step 2 - Solve. $c=\dfrac{8\pm\sqrt{864}}{50}=\dfrac{4\pm6\sqrt6}{25}$. In $\left[0,\frac\pi4\right]$ we need $c\ge\frac{1}{\sqrt2}$, so $c=\dfrac{4+6\sqrt6}{25}\approx0.748$. Step 3 - Match the options. $(4+6\sqrt6)(3\sqrt6-2)=12\sqrt6-8+108-12\sqrt6=100$, so $$\frac{4+6\sqrt6}{25}=\frac{4}{3\sqrt6-2}.$$ Answer: (B)