# im really quite confused about this problem so can someone help me in this one?? Solve for x of `(4 sin^2 x -1) (cos x - 2) = 0` where 0 ≤ x ≤ 2π .

*print*Print*list*Cite

`(4sin^2x-1)(cos x - 2)=0`

To solve for x, factor 4sin^2x - 1.

`(2sinx-1)(2sinx+1)(cosx - 2)=0`

Then, set each factor equal to zero and solve for x.

For the first factor:

`2sin x - 1 = 0`

`2sinx -1+1=0+1`

`2sinx=1`

`(2sinx)/2=1/2`

`sinx=1/2`

Refer to Unit Circle Chart to determine the value of the angle x.

`x=pi/6 and (5pi)/6`

For the second factor:

`2sinx+1=0`

`2sinx+1-1=0-1`

`2sinx=-1`

`(2sinx)/2=-1/2`

`sinx=-1/2`

Again, refer to Unit Circle Chart to determine the angle.

`x=(7pi)/6 and (11pi)/6`

And the third factor:

`cos x-2=0`

`cosx-2+2=0+2`

`cosx=2 ` (Invalid equation)

For the third factor apply the properties of cosine. Its range is -1 `<=` cosx `<=` 1. Since 2 does not fall within this interval, the third factor has no solution.

**Hence, the solutions to the equation `(4sin^2x-1)(cosx-2)=0` **

**are `x= pi/6, (5pi)/6, (7pi)/6 and (11pi)/6` .**