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If secxcos5x+1=0, where 0 ltxlt=pi/2. fi...

If `secxcos5x+1=0`, where `0 ltxlt=pi/2`. find the value of `x` .

A

`(pi)/(6)`

B

`(pi)/(4)`

C

`(pi)/(2)`

D

Both (a) and (b)

Text Solution

Verified by Experts

The correct Answer is:
D
Given , `secxcos5x+1=0`
`rArr(1)/(cosx)*cos5x+(1)/(1)=0`
`rArr(cos5x+cosx)/(cosx)=0`
`rArrcos5x+cosx=0,cosxne0`
`rArr2"cos"(5x+x)/(2)"cos"(5x-x)/(2)=0`
`rArr2cos3xcos2x=0`
`cos3x=0orcos2x=0`
`rArr3x=(2n+1)(pi)/(2)or2x=(2n+1)(pi)/(2)`
`rArrx=(2n+1)(pi)/(6)orx=(2n+1)(pi)/(4)`
If n = 0 , then `x=(pi)/(6)orx=(pi)/(4)`
If n = 1 , then
`x=3xx(pi)/(6)=(pi)/(2)orx=(3pi)/(4)gt(pi)/(2)`
Therefore ,the values of x are `(pi)/(6),(pi)/(4)`.
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