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 .
`(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`
Refer to Unit Circle Chart to determine the value of the angle x.
`x=pi/6 and (5pi)/6`
For the second factor:
Again, refer to Unit Circle Chart to determine the angle.
`x=(7pi)/6 and (11pi)/6`
And the third factor:
`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` .