JEE MAIN - Mathematics (2024 - 5th April Morning Shift - No. 17)

Suppose $$\theta \in\left[0, \frac{\pi}{4}\right]$$ is a solution of $$4 \cos \theta-3 \sin \theta=1$$. Then $$\cos \theta$$ is equal to :
$$\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}$$
$$\frac{4}{(3 \sqrt{6}+2)}$$
$$\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}$$
$$\frac{4}{(3 \sqrt{6}-2)}$$

Explanation

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

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

Comments (0)

Advertisement