1.

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

Answer» Given that, `" "secxcos5x+1=0`
`" "(cos5x)/(cosx)+1=0rArrcos5x+cosx=0`
`rArr2cos((5x+x)/(2))*cos((5x-x)/(2))=0` `" "[because cosx+cosy=2cos""(x+y)/(2)*cos""(x-y)/(2)]`
`rArr" "2cos3x*cos2x=0`
`rArr" "cos3x=0orcos2x=0`
`rArr " "cos3x=cos""(pi)/(2) or cos2x=cos""(pi) /( 2)`
`therefore" "3x=(pi)/(2)rArr2x=(pi)/(2)`
and `" "x=(pi)/(6)rArrx=(pi)/(4)`
Hence, the solutions are `(pi)/(2), (pi)/(4) and (pi)/(6)`.


Discussion

No Comment Found

Related InterviewSolutions