For theta in left[-frac{pi}{2}, frac{pi}{2}ri | vedclass

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Question

(B) Given $2 \cos 	heta + \sin 	heta = 1$. Since $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$,$\cos 	heta \ge 0$. Let $\cos 	heta = x$

Vedclass · Accepted Answer

$6, 2$

Vedclass · Answer

(B) Given $2 \cos 	heta + \sin 	heta = 1$. Since $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$,$\cos 	heta \ge 0$. Let $\cos 	heta = x$ and $\sin 	heta = y$. Then $2x + y = 1 \implies y = 1 - 2x$. Using $x^2 + y^2 = 1$,we have $x^2 + (1 - 2x)^2 = 1$. $x^2 + 1 - 4x + 4x^2 = 1 \implies 5x^2 - 4x = 0$. So,$x(5x - 4) = 0$,which gives $x = 0$ or $x = \frac{4}{5}$. Case $1$: If $x = 0$,then $\cos 	heta = 0$. Since $	heta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}ight]$,$	heta = \frac{\pi}{2}$ or $-\frac{\pi}{2}$. If $	heta = \frac{\pi}{2}$,$\sin 	heta = 1$,then $k = 7(0) + 6(1) = 6$. If $	heta = -\frac{\pi}{2}$,$\sin 	heta = -1$,then $k = 7(0) + 6(-1) = -6$. Case $2$: If $x = \frac{4}{5}$,then $\cos 	heta = \frac{4}{5}$. Since $y = 1 - 2x$,$y = 1 - 2(\frac{4}{5}) = 1 - \frac{8}{5} = -\frac{3}{5}$. Then $k = 7(\frac{4}{5}) + 6(-\frac{3}{5}) = \frac{28}{5} - \frac{18}{5} = \frac{10}{5} = 2$. The possible values of $k$ are $6, -6, 2$.

For $\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$,if $2 \cos \theta + \sin \theta = 1$ and $7 \cos \theta + 6 \sin \theta = k$,then the possible values of $k$ are:

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