1.

Find number of solution of the equation `2 sin x+5 sin^(2) x+8sin^(3)x+... oo=1` for `x in [0, 2pi]`.

Answer» Correct Answer - Two solutions
We have
`2 sin x+5 sin^(2)x+8sin^(3)x+...=1` ...(1)
`:. 2sin^(2)x+5 sin^(3) x+...=sin x` ...(2)
Substracting (2) from (1), we get
`2 sin x+3(sin^(2)x+sin^(3)x+...)=1-sin x`
`:. 2sin x+ (3 sin^(2) x)/(1- sin x)=1-sin x`
`rArr 2 sin x-2 sin^(2)x+3 sin^(2)x=1-2 sin x+sin^(2)x`
`rArr sin x=1/4`
So there will be two solutions.


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