1.

The equation `2cos^2x/2sin^2x=x^2+x^(-2);0`

Answer» `(2cos^2x/2)sin^2x=x^2+1/x^2`
LHS
`0ltxltpi/2`
`cos0>cosx>=cospi/2`
`1>cosx>=0`
`2>1+cosx>=1`
`1<1+cosx<2`
`-1<=sinx<=1`
`0<=(1+cosx)(sin^2x)<2`
`0<=LHS<=2`
RHS
`x^2+1/x^2=(x-1/x)^2+2`
`0<=(x-1/x)^2`2<=(x-1/x)^2+2`2<=RHS<=oo`.


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