1.

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

Answer» Correct Answer - Six solutions
We have
`cot^(2) (sin x+3)=1="cot"^(2) pi/4`
`rArr sin x+3=n pi pm pi/4`
Now, `2 le sin x +3 le 4`
`:. Sin x+3=pi-pi/4`
or `sin x+3=pi+pi/4`
`:. Sin x=(3pi)/4-3` ...(1)
or `sin x=(5pi)/4-3` ...(2)
For (1), we have two values of x in interval `(pi, 2pi)`.
For (2), we have two values of x in each of the intervals `(0, pi), (2pi, 3pi)`.
So, there are total six solutions.


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