1.

The number of values of `x`in the interval `0,5pi`satisfying the equation `3sin^2x-7sinx+2=0`is`0`(b) `5`(c) `6`(d) `10`

Answer» `3sin^2x-7sinx+2 = 0`
`=>3sin^2x-6sinx-sinx+2 = 0`
`=>(sinx-2)(3sinx-1) = 0`
`=>sinx = 2 or sinx = 1/3`
But, `sin x` can not be more than `1`.
`:. sinx = 1/3`
It means, value of `sinx` is positive.
It means, `x` lies in first and second quadrant.
So, there are `2` values from `0` to `2pi`.
So, from `0` to `5pi`, there will be `5` values for `x` that will satisfy the given equation.


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