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secx.cos5x+1=0...

`secx.cos5x+1=0`

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If secxcos5x+1=0 , where 0ltxle(pi)/(2) , then find the value of x.

If secxcos5x+1=0 , where 0ltxle(pi)/(2) , then find the value of x.

If secxcos5x+1=0,"where "0ltxle(pi)/(2) , then find the value of x.

The number of solutions of secxcos5x+1=0 in the interval [0,2pi] is

The number of solutions of secxcos5x+1=0 in the interval [0,2pi] is

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lim_(x->0)(log(1+x+x^2)+"log"(1-x+x^2))/(secx-cosx)=

lim_(xto0) (log(1+x+x^(2))+log(1-x+x^(2)))/(secx-cosx)=

lim_(xto0) (log(1+x+x^(2))+log(1-x+x^(2)))/(secx-cosx)=

The value of cos(pi+x)sin(pi-x) secx.coses x is