cos(2x + 7) = a(2 - sin x) can have a real so | vedclass

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(B) Step $1$. Rewrite the equation using the identity $\cos(2x) = 1 - 2\sin^2 x$: $1 - 2\sin^2 x = 2a - a\sin x$ $2\sin^2 x - a\sin x + 2a - 1 = 0$ Wa

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$a \in [2, 6]$

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(B) Step $1$. Rewrite the equation using the identity $\cos(2x) = 1 - 2\sin^2 x$: $1 - 2\sin^2 x = 2a - a\sin x$ $2\sin^2 x - a\sin x + 2a - 1 = 0$ Wait, let us re-evaluate the given equation $\cos(2x + 7) = a(2 - \sin x)$. Actually, the standard form is $2\sin^2 x - a\sin x + (2a - 8) = 0$. Step $2$. Solve for $\sin x$ using the quadratic formula: $\sin x = \frac{a \pm \sqrt{a^2 - 4(2)(2a - 8)}}{2(2)} = \frac{a \pm \sqrt{a^2 - 16a + 64}}{4} = \frac{a \pm \sqrt{(a - 8)^2}}{4} = \frac{a \pm (a - 8)}{4}$. This gives two roots: $\sin x = \frac{a + a - 8}{4} = \frac{2a - 8}{4} = \frac{a - 4}{2}$ $\sin x = \frac{a - a + 8}{4} = \frac{8}{4} = 2$ (Not possible as $\sin x \leq 1$). Step $3$. Since $-1 \leq \sin x \leq 1$, we have: $-1 \leq \frac{a - 4}{2} \leq 1$ $-2 \leq a - 4 \leq 2$ $2 \leq a \leq 6$. Thus, $a \in [2, 6]$.

$\cos(2x + 7) = a(2 - \sin x)$ can have a real solution for

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