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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.

A

`(pi)/(5),(pi)/(4)`

B

`(pi)/(5)`

C

`(pi)/(4)`

D

None of these

Text Solution

Verified by Experts

The correct Answer is:
C
`secxcos5x=-1`
`rArrcos5x=-cosx`
`rArr5x=2npi+-(pi-x)`
`rArrx=((2n+1)pi)/(6)or((2n-1)pi)/(4)`
The possible values of x which lies in the interval `(0,2pi)` are
`(pi)/(6),(pi)/(4),(pi)/(2),(3pi)/(4),(5pi)/(6),(5pi)/(4),(7pi)/(6),(7pi)/(4),(9pi)/(6)and(11pi)/(6)`.
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