If the equation tan^4x - 2sec^2x + [a]^2 = 0 | vedclass

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(C) Given the equation $	an^4x - 2\sec^2x + [a]^2 = 0$. Using the identity $\sec^2x = 1 + 	an^2x$,we get: $	an^4x - 2(1 + 	an^2x) + [a]^2 = 0$ $

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$[-1, 2)$

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(C) Given the equation $	an^4x - 2\sec^2x + [a]^2 = 0$. Using the identity $\sec^2x = 1 + 	an^2x$,we get: $	an^4x - 2(1 + 	an^2x) + [a]^2 = 0$ $	an^4x - 2	an^2x - 2 + [a]^2 = 0$ $(	an^2x - 1)^2 - 1 - 2 + [a]^2 = 0$ $(	an^2x - 1)^2 = 3 - [a]^2$ Since $(	an^2x - 1)^2 \geq 0$,we must have $3 - [a]^2 \geq 0$,which implies $[a]^2 \leq 3$. Thus,$[a] \in [-\sqrt{3}, \sqrt{3}]$. Since $[a]$ is an integer,$[a] \in \{-1, 0, 1\}$. If $[a] = -1$,then $-1 \leq a If $[a] = 0$,then $0 \leq a If $[a] = 1$,then $1 \leq a Combining these,the range of $a$ is $[-1, 2)$.

If the equation $\tan^4x - 2\sec^2x + [a]^2 = 0$ has at least one solution,then the complete range of $a$ (where $a \in R$) is:
(Note: $[k]$ denotes the greatest integer less than or equal to $k$)

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If the equation $\tan^4x - 2\sec^2x + [a]^2 = 0$ has at least one solution,then the complete range of $a$ (where $a \in R$) is: (Note: $[k]$ denotes the greatest integer less than or equal to $k$)