If 2costheta + sintheta = 1,then the value of | vedclass

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

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

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(D) Given $2\cos	heta + \sin	heta = 1$,we can write $2\cos	heta = 1 - \sin	heta$.Squaring both sides,we get $(2\cos	heta)^2 = (1 - \sin	heta)^2$.$4\cos^2	heta = 1 - 2\sin	heta + \sin^2	heta$.Using $\cos^2	heta = 1 - \sin^2	heta$,we have $4(1 - \sin^2	heta) = 1 - 2\sin	heta + \sin^2	heta$.$4 - 4\sin^2	heta = 1 - 2\sin	heta + \sin^2	heta$.$5\sin^2	heta - 2\sin	heta - 3 = 0$.Factoring the quadratic equation: $(5\sin	heta + 3)(\sin	heta - 1) = 0$.This gives $\sin	heta = 1$ or $\sin	heta = -\frac{3}{5}$.Case $1$: If $\sin	heta = 1$,then $\cos	heta = 0$. Substituting into the expression $E = 4\cos	heta + 3\sin	heta$,we get $E = 4(0) + 3(1) = 3$.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 $2\cos	heta + \sin	heta = 1$,we have $2\cos	heta = 1 - (-\frac{3}{5}) = \frac{8}{5}$,which implies $\cos	heta = \frac{4}{5}$.Substituting into $E$,we get $E = 4(\frac{4}{5}) + 3(-\frac{3}{5}) = \frac{16}{5} - \frac{9}{5} = \frac{7}{5}$.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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