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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` |
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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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