If 2cos⁡θ+sin⁡θ=1,(θ≠(4k+1)π/2k∈I ) then 7cos⁡θ+6sin⁡θ is equal to

2cos⁡θ+sin⁡θ=1⇒4cos2⁡θ=(1−sin⁡θ)2⇒41−sin2⁡θ=(1−sin⁡θ)2⇒4(1+sin⁡θ)=1−sin⁡θ⇒sin⁡θ=−3/5As 2cos⁡θ+sin⁡θ=1, we get cos⁡θ=4/5. Thus 7cos⁡θ+6sin⁡θ=745+6−35=2

If 2cos⁡θ+sin⁡θ=1,(θ≠(4k+1)π/2k∈I ) then 7cos⁡θ+6sin⁡θ is equal to

2cos⁡θ+sin⁡θ=1⇒4cos2⁡θ=(1−sin⁡θ)2⇒41−sin2⁡θ=(1−sin⁡θ)2⇒4(1+sin⁡θ)=1−sin⁡θ⇒sin⁡θ=−3/5As 2cos⁡θ+sin⁡θ=1, we get cos⁡θ=4/5. Thus 7cos⁡θ+6sin⁡θ=745+6−35=2

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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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Detailed Solution

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

$\begin{array}{l} \Rightarrow 4 (1 - \sin^{2} θ) = (1 - \sin θ)^{2} \\ \Rightarrow 4 (1 + \sin θ) = 1 - \sin θ \\ \Rightarrow \sin θ = - 3 / 5 \end{array}$