The set of all values of lambda for which the | vedclass

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(D) Given equation: $\lambda = \cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x$Convert all terms to $\cos x$:$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^

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$[-\frac{3}{2}, -1]$

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(D) Given equation: $\lambda = \cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x$Convert all terms to $\cos x$:$\lambda = (2 \cos ^2 x - 1)^2 - 2(1 - \cos ^2 x)^2 - 2 \cos ^2 x$Expand the terms:$\lambda = (4 \cos ^4 x - 4 \cos ^2 x + 1) - 2(1 - 2 \cos ^2 x + \cos ^4 x) - 2 \cos ^2 x$Simplify:$\lambda = 4 \cos ^4 x - 4 \cos ^2 x + 1 - 2 + 4 \cos ^2 x - 2 \cos ^4 x - 2 \cos ^2 x$$\lambda = 2 \cos ^4 x - 2 \cos ^2 x - 1$Let $t = \cos ^2 x$,where $t \in [0, 1]$:$f(t) = 2t^2 - 2t - 1$Find the range of $f(t)$ for $t \in [0, 1]$:$f'(t) = 4t - 2$. Setting $f'(t) = 0$ gives $t = \frac{1}{2}$.$f(0) = -1$$f(1) = 2 - 2 - 1 = -1$$f(\frac{1}{2}) = 2(\frac{1}{4}) - 2(\frac{1}{2}) - 1 = \frac{1}{2} - 1 - 1 = -\frac{3}{2}$Thus,the range of $\lambda$ is $[-\frac{3}{2}, -1]$.

The set of all values of $\lambda$ for which the equation $\cos ^2 2x - 2 \sin ^4 x - 2 \cos ^2 x = \lambda$ has a solution is:

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