1.

The number of values of x in the interval [0,5pi] satisfying the equation. 3sin^(2)x -7sinx + 2=0 is-

Answer»

0
5
6
10

Solution :`3sin^(2)x-7sinx+2=0`
`rArr (3sinx-1)(sinx-2)=0`
`therefore sinx ne 2`
`rArr sinx=1/3 =sinalpha` (SAY)
Where `alpha` is the least positive VALUE of x
such that `sinalpha=1/3`

Clearly, `0 lt alpha lt pi/2`. We get the solution,
`x=alpha, pi-alpha, 2pi+alpha, 4PI + alpha` and `5pi- alpha`
Hence total six VALUES in `[0,5pi]` Ans.


Discussion

No Comment Found