1.

Find the number of distinct solutions of secx +tanx = sqrt(3), where 0 le xle3pi.

Answer»

Solution :Here, `secx +tanx=sqrt(3) rArr 1+sinx = sqrt(3)cosx`
or `sqrt(3)cosx-sinx=1`
`sqrt(3)cosx-sinx=1`
DIVIDING both sides by `sqrt(a^(2)+b^(2))`. i.e., `sqrt(4)=2`, we get
`rArr sqrt(3)/2 cosx-1/xsinx=1/2`
`rArr cospi/6cosx-sinpi/6 sinx=1/2 rArr cos(x+pi/6) = 1/2`
As `0lexle3pi`
`pi/6 le x +pi/6 le3pi+pi/6`
`rArr x+pi/6=pi/3, (5pi)/3, (7pi)/(3) rArr x=pi/6, (3pi)/(2), (13pi)/(6)`
But at `x=(3pi)/(2)`,`tanx` and `secx` is not defined
`THEREFORE` Total NUMBER of solutions are 2.


Discussion

No Comment Found

Related InterviewSolutions