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JEE Main 2018
Trigonometry
Trigonometric Ratio and Identites
Hard

Question

Suppose θ[0,π4]\theta \in\left[0, \frac{\pi}{4}\right] is a solution of 4cosθ3sinθ=14 \cos \theta-3 \sin \theta=1. Then cosθ\cos \theta is equal to :

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Solution

4cosθ3sinθ=14cosθ1=3sinθ16cos2θ+18cosθ=9(1cos2θ)25cos2θ8cosθ8=0cosθ=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} =4±25425 As θ[0,π4]cosθ=4+6625=4362\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}

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