QJEEM24Ma3ff2: JEE Main 2024 Mathematics - Class 11 JEEM24Ma3ff2 Solution: Suppose is a solution of . Then is equal to :

$\begin{aligned}& 4 \cos \theta-3 \sin \theta=1 \\& 4 \cos \theta-1=3 \sin \theta \\& 16 \cos ^2 \theta+1-8 \cos \theta=9\left(1-\cos ^2 \theta\right) \\& \Rightarrow 25 \cos ^2 \theta-8 \cos \theta-8=0 \\& \Rightarrow \cos \theta=\frac{8 \pm \sqrt{64+4 \times 25 \times 8}}{2.25} \\& =\frac{8 \pm 4 \sqrt{4+50}}{2.25}\end{aligned}$

$\begin{aligned}& =\frac{4 \pm 2 \sqrt{54}}{25} \\& \text { As } \theta \in\left[0, \frac{\pi}{4}\right] \\& \Rightarrow \cos \theta=\frac{4+6 \sqrt{6}}{25}=\frac{4}{3 \sqrt{6}-2}\end{aligned}$