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

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Question

$\left(	an ^{2} x-1ight)^{2}=3-[a]^{2}$

$	ext { Hence, } 3-[a]^{2} \geq 0 \Rightarrow[a] \in[-\sqrt{3}, \sqrt{3}]$

$	herefore[a]=-1,0,1$

Vedclass · Accepted Answer

$[-1, 2)$

Vedclass · Answer

$\left(	an ^{2} x-1ight)^{2}=3-[a]^{2}$

$	ext { Hence, } 3-[a]^{2} \geq 0 \Rightarrow[a] \in[-\sqrt{3}, \sqrt{3}]$

$	herefore[a]=-1,0,1$

$\Rightarrow \mathrm{a} \in[-1,2)$

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

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