1.

Find the number of solutions of `tanx + secx = 2cosx` in `[0,2pi]`

Answer» Here, `tanx + secx = 2cosx rArr sinx +1=2cos^(2)x`
`rArr 2sin^(2)x+sinx-1=0 rArr sinx=1/2, -1`
But `sinx=-1 rArr x=(3pi)/(2)` for which `tanx+secx =2 cosx` is not defined.
Thus, `sinx=1/2 rArr x=pi/6, (5pi)/(6)`
`rArr` number of solutions of `tanx+secx = 2cos x` is 2. Ans.


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