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

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

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$2$

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(D) Given $2 \cos 	heta + \sin 	heta = 1$. Squaring both sides,we get $(2 \cos 	heta + \sin 	heta)^2 = 1^2$. $4 \cos^2 	heta + \sin^2 	heta + 4 \sin 	heta \cos 	heta = 1$. Since $\sin^2 	heta = 1 - \cos^2 	heta$,we have $4 \cos^2 	heta + (1 - \cos^2 	heta) + 4 \sin 	heta \cos 	heta = 1$. $3 \cos^2 	heta + 4 \sin 	heta \cos 	heta = 0$. $\cos 	heta (3 \cos 	heta + 4 \sin 	heta) = 0$. Since $	heta 
eq \frac{\pi}{2}$,$\cos 	heta 
eq 0$,so $3 \cos 	heta + 4 \sin 	heta = 0$,which implies $	an 	heta = -\frac{3}{4}$. Using $	an 	heta = -\frac{3}{4}$,we can form a right triangle with opposite side $3$ and adjacent side $4$. Since $	an 	heta$ is negative,$	heta$ is in the second or fourth quadrant. Given $2 \cos 	heta + \sin 	heta = 1$,if $\cos 	heta = \frac{4}{5}$ and $\sin 	heta = -\frac{3}{5}$,then $2(\frac{4}{5}) + (-\frac{3}{5}) = \frac{8-3}{5} = 1$ (Satisfied). If $\cos 	heta = -\frac{4}{5}$ and $\sin 	heta = \frac{3}{5}$,then $2(-\frac{4}{5}) + \frac{3}{5} = -1 
eq 1$. Thus,$\cos 	heta = \frac{4}{5}$ and $\sin 	heta = -\frac{3}{5}$. Therefore,$7 \cos 	heta + 6 \sin 	heta = 7(\frac{4}{5}) + 6(-\frac{3}{5}) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$.

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

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