If 2 cos theta + sin theta = 1,then the value | vedclass

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(D) Given $2 \cos 	heta + \sin 	heta = 1$,we have $2 \cos 	heta = 1 - \sin 	heta$. Squaring both sides: $4 \cos^2 	heta = (1 - \sin 	heta)^2$. U

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Both $(a)$ and $(c)$

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(D) Given $2 \cos 	heta + \sin 	heta = 1$,we have $2 \cos 	heta = 1 - \sin 	heta$. Squaring both sides: $4 \cos^2 	heta = (1 - \sin 	heta)^2$. Using $\cos^2 	heta = 1 - \sin^2 	heta$: $4(1 - \sin^2 	heta) = 1 - 2 \sin 	heta + \sin^2 	heta$. $4 - 4 \sin^2 	heta = 1 - 2 \sin 	heta + \sin^2 	heta \Rightarrow 5 \sin^2 	heta - 2 \sin 	heta - 3 = 0$. Factoring the quadratic: $(5 \sin 	heta + 3)(\sin 	heta - 1) = 0$. So,$\sin 	heta = 1$ or $\sin 	heta = -\frac{3}{5}$. Case $1$: If $\sin 	heta = 1$,then $\cos 	heta = 0$. Then $4 \cos 	heta + 3 \sin 	heta = 4(0) + 3(1) = 3$. Case $2$: If $\sin 	heta = -\frac{3}{5}$,then $\cos^2 	heta = 1 - (-\frac{3}{5})^2 = 1 - \frac{9}{25} = \frac{16}{25}$,so $\cos 	heta = \pm \frac{4}{5}$. Since $2 \cos 	heta = 1 - \sin 	heta = 1 - (-\frac{3}{5}) = \frac{8}{5}$,we must have $\cos 	heta = \frac{4}{5}$. Then $4 \cos 	heta + 3 \sin 	heta = 4(\frac{4}{5}) + 3(-\frac{3}{5}) = \frac{16}{5} - \frac{9}{5} = \frac{7}{5}$. Thus,the possible values are $3$ and $\frac{7}{5}$.

If $2 \cos \theta + \sin \theta = 1$,then the value of $4 \cos \theta + 3 \sin \theta$ is equal to

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