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The equation $\cos^{8} x + {bcos}^{4} x + 1 = 0$ will have a solution if $b$ belongs to

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$(- \infty, 2]$

$[2, \infty)$

$(- \infty, - 2]$

answer is C.

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Detailed Solution

$\cos^{8} x + b \cos^{4} x + 1 = 0 \cos^{4} x = \frac{- b \pm \sqrt{b^{2} - 4}}{2} \geq 0 b^{2} - 4 \geq 0 \Rightarrow (b - 2) (b + 2) \geq 0 b \in (- \infty, - 2) \cup [2, \infty] \cos^{4} x > 0 \Rightarrow b \in (- \infty, - 2]$

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