JEE Main Previous Year Trigonometry Questions and Solutions - Solution of Triangles Trigonometric equations

Question

Suppose $θ \in [0, \frac{π}{4}]$ is a solution of $4 \cos θ - 3 \sin θ = 1$ . Then $\cos θ$ is equal to : [2024]

Smart Achievers · Accepted Answer

(2)

We have, $4 \cos θ - 3 \sin θ = 1$

$\Rightarrow 4 \cos θ - 1 = 3 \sin θ \Rightarrow {(4 \cos θ - 1)}^{2} = {(3 \sin θ)}^{2}$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 \sin^{2} θ$

$\Rightarrow 16 \cos^{2} θ + 1 - 8 \cos θ = 9 (1 - \cos^{2} θ)$

$\Rightarrow 25 \cos^{2} θ - 8 \cos θ - 8 = 0$

$\Rightarrow \cos θ = \frac{8 \pm \sqrt{64 + 800}}{50}$

$= \frac{8 \pm \sqrt{864}}{50} = \frac{8 \pm 12 \sqrt{6}}{50} = \frac{4 \pm 6 \sqrt{6}}{25}$

Since, $θ \in [0, π / 4]$ so $\cos θ = \frac{4 + 6 \sqrt{6}}{25}$

$= \frac{(4 - 6 \sqrt{6}) (4 + 6 \sqrt{6})}{25 (4 - 6 \sqrt{6})} = \frac{- 200}{25 \times 2 (2 - 3 \sqrt{6})} = \frac{4}{3 \sqrt{6} - 2}$

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