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If secxcos5x+1=0 , where 0ltxle(pi)/(2)...

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

Text Solution

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`secx*cosx+1=0`
`(cos5x)/cosx=-1`
`cos5x=-cosx`
`cos5x+cosx=0`
`2cos((5x+x)/2)*cos((5x-x)/2)=0`
`2 cos3x*cos2x=0`
`cos3x=0`
`3x=pi/2,3/2pi`
...
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