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Trigonometric equations

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Question

If $2 \cos θ + \sin θ = 1, (θ \neq (4 k + 1) π / 2$ $k \in I)$ then $7 \cos θ + 6 \sin θ$ is equal to

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Solution

$2 \cos θ + \sin θ = 1 \Rightarrow 4 \cos^{2} θ = (1 - \sin θ)^{2}$
$\begin{array}{l} \Rightarrow 4 (1 - \sin^{2} θ) = (1 - \sin θ)^{2} \\ \Rightarrow 4 (1 + \sin θ) = 1 - \sin θ \\ \Rightarrow \sin θ = - 3 / 5 \end{array}$
As $2 \cos θ + \sin θ = 1,$ we get $\cos θ = 4 / 5 .$
Thus $7 \cos θ + 6 \sin θ = 7 (\frac{4}{5}) + 6 (\frac{- 3}{5}) = 2$

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