If sin theta + 2 cos theta = 1 and theta belo | vedclass

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(B) Given $\sin 	heta + 2 \cos 	heta = 1$. Rearranging,we get $\sin 	heta = 1 - 2 \cos 	heta$. Squaring both sides: $\sin^2 	heta = (1 - 2 \cos \

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$2$

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(B) Given $\sin 	heta + 2 \cos 	heta = 1$. Rearranging,we get $\sin 	heta = 1 - 2 \cos 	heta$. Squaring both sides: $\sin^2 	heta = (1 - 2 \cos 	heta)^2$. Using $\sin^2 	heta = 1 - \cos^2 	heta$,we have $1 - \cos^2 	heta = 1 - 4 \cos 	heta + 4 \cos^2 	heta$. Simplifying: $5 \cos^2 	heta - 4 \cos 	heta = 0$. Thus,$\cos 	heta (5 \cos 	heta - 4) = 0$. Since $	heta$ is in the $4^{	ext{th}}$ quadrant and not on the axes,$\cos 	heta 
eq 0$. Therefore,$\cos 	heta = \frac{4}{5}$. Substituting back: $\sin 	heta = 1 - 2(\frac{4}{5}) = 1 - \frac{8}{5} = -\frac{3}{5}$. Now,calculate $7 \cos 	heta + 6 \sin 	heta = 7(\frac{4}{5}) + 6(-\frac{3}{5}) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$.

If $\sin \theta + 2 \cos \theta = 1$ and $\theta$ belongs to the $4^{\text{th}}$ quadrant (not lying on the coordinate axes),then $7 \cos \theta + 6 \sin \theta = $

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