JEE Main 2018TrigonometryTrigonometric Ratio and IdentitesHardQuestionSuppose θ∈[0,π4]\theta \in\left[0, \frac{\pi}{4}\right]θ∈[0,4π] is a solution of 4cosθ−3sinθ=14 \cos \theta-3 \sin \theta=14cosθ−3sinθ=1. Then cosθ\cos \thetacosθ is equal to :OptionsA6−6(36−2)\frac{6-\sqrt{6}}{(3 \sqrt{6}-2)}(36−2)6−6B4(36+2)\frac{4}{(3 \sqrt{6}+2)}(36+2)4C6+6(36+2)\frac{6+\sqrt{6}}{(3 \sqrt{6}+2)}(36+2)6+6D4(36−2)\frac{4}{(3 \sqrt{6}-2)}(36−2)4Check AnswerHide SolutionSolution4cosθ−3sinθ=14cosθ−1=3sinθ16cos2θ+1−8cosθ=9(1−cos2θ)⇒25cos2θ−8cosθ−8=0⇒cosθ=8±64+4×25×82.25=8±44+502.25\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}4cosθ−3sinθ=14cosθ−1=3sinθ16cos2θ+1−8cosθ=9(1−cos2θ)⇒25cos2θ−8cosθ−8=0⇒cosθ=2.258±64+4×25×8=2.258±44+50 =4±25425 As θ∈[0,π4]⇒cosθ=4+6625=436−2\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}=254±254 As θ∈[0,4π]⇒cosθ=254+66=36−24