If 2,cos,theta + sin, theta , = 1 left( | vedclass

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Question

Given $2 \cos 	heta+\sin 	heta=1$

Squaring both sides, we get $(2 \cos 	heta+\sin 	heta)^{2}=1^{2}$

$\Rightarrow 4 \cos ^{2} 	heta+\sin ^{2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

Given $2 \cos 	heta+\sin 	heta=1$

Squaring both sides, we get $(2 \cos 	heta+\sin 	heta)^{2}=1^{2}$

$\Rightarrow 4 \cos ^{2} 	heta+\sin ^{2} 	heta+4 \sin 	heta \cos 	heta=1$

$\Rightarrow 3 \cos ^{2} 	heta+\left(\cos ^{2} 	heta+\sin ^{2} 	hetaight)+4 \sin 	heta \cos 	heta$ $=1$

$\Rightarrow 3 \cos ^{2} 	heta+1+4 \sin 	heta \cos 	heta=1$

$\Rightarrow 3 \cos ^{2} 	heta+4 \sin 	heta \cos 	heta=0$

$\Rightarrow \cos 	heta(3 \cos 	heta+4 \sin 	heta)=0$

$\Rightarrow 3 \cos 	heta+4 \sin 	heta=0$

$\Rightarrow 3 \cos 	heta=-4 \sin 	heta$

$\Rightarrow \frac{-3}{4}=	an 	heta=\sqrt{\sec ^{2} 	heta-1}=\frac{-3}{4}$

$(\because 	an 	heta=\sqrt{\sec ^{2} 	heta-1})$

$\Rightarrow \sec ^{2} 	heta-1=$ $\left(\frac{-3}{4}ight)^{2}=\frac{9}{16}$

$\Rightarrow \sec ^{2} 	heta=\frac{9}{16}+1=$ $\frac{25}{16} \Rightarrow \sec 	heta=\frac{5}{4}$

or ${\cos 	heta=\frac{4}{5}}$    ....... $(1)$

Now, $\sin ^{2} 	heta+\cos ^{2} 	heta=1$

$\Rightarrow \sin ^{2} 	heta+\left(\frac{4}{5}ight)^{2}=1$

$\sin ^{2} 	heta+\frac{4}{5}=1 \Rightarrow \sin ^{2} 	heta=1-\frac{16}{25}=\frac{9}{25}$

$\sin 	heta=\pm \frac{3}{5}$     ........ $(2)$

Taking $\quad\left(\sin 	heta=-\frac{3}{5}ight) \quad$ because $\left(\sin 	heta=\frac{3}{5}ight)$ cannot satisfy the given equation.

Therefore; $7 \cos 	heta+6 \sin 	heta$

$=7 	imes \frac{4}{5}+6 	imes \frac{3}{5}=\frac{28}{5}-\frac{18}{5}=2$

If $2\,cos\,\theta + sin\, \theta \, = 1$ $\left( {\theta \ne \frac{\pi }{2}} \right)$ , then $7\, cos\,\theta + 6\, sin\, \theta $ is equal to

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