Question

`cosx(2sinx + sqrt3)(-sqrt2 cosx + 1) = 0`??

Answer 1

Maybe you meant "how do I solve this equation". By the zero-product property, one or all of these factors can be equal to `0`.

`cosx(2sinx+sqrt3)( -sqrt2cosx+1) = 0`

• When `cosx = 0`, `x = pi/2, (3pi)/2, . . .`, or:

`color(blue)(x = pi/2 pm npi)`

• When `2sinx + sqrt3 = 0`, `sinx = -sqrt3/2`. That can occur for `240^@`, `600^@`, `960^@`, etc., and `300^@`, `660^@`, `1020^@`, etc. Or:

`color(blue)(x = (4pi)/3 pm 2npi)`, where `n` is an integer.
`color(blue)(x = (5pi)/3 pm 2npi)`, where `n` is an integer.

• When `1 - sqrt2cosx = 0`, `cosx = 1/sqrt2*(sqrt2/sqrt2) = sqrt2/2`. This occurs when `x = 45^@, 315^@, 405^@, 675^@, . . .`. So `x = pi/4, (7pi)/4, (9pi)/4, (15pi)/4, . . .`, or:

`color(blue)(x = 2npi pm pi/4)`, where `n` is an integer.

Note that written this way, this already includes both `pi/4` and `(7pi)/4` for `0 < x < 2pi` and other corresponding angles in subsequent cycles. For example:

`n = -1 -> x = -(9pi)/4, -(7pi)/4`
`n = 0 -> x = -pi/4, pi/4`
`n = 1 -> x = (7pi)/4, (9pi)/4`
`n = 2 -> x = (15pi)/4, (17pi)/4`

---

If I took this at face value, I honestly don't think this is equal to `0`.

http://www.wolframalpha.com/input/?i=Is+cosx%282sinx%2Bsqrt3%29%28+-sqrt2cosx%2B1%29+%3D+0

If we tried, we would get:

`cosx(2sinx+sqrt3)( -sqrt2cosx+1) stackrel(?)(=) 0`

`(2sinxcosx+sqrt3cosx)( -sqrt2cosx+1) stackrel(?)(=) 0`

`-2sqrt2sinxcos^2x + 2sinxcosx - sqrt6cos^2x + sqrt3cosx stackrel(?)(=) 0`

`2sinxcosx + sqrt3cosx stackrel(?)(=) 2sqrt2sinxcos^2x + sqrt6cos^2x`

`(2sinx + sqrt3)cosx stackrel(?)(=) (2sqrt2sinx + sqrt6)cos^2x`

`cancel((2sinx + sqrt3))cosx stackrel(?)(=) sqrt2cancel((2sinx + sqrt3))cos^2x`

`cosx stackrel(?)(=) sqrt2cos^2x`

`1 stackrel(?)(=) sqrt2cosx`

`1/sqrt2 = sqrt2/2 ne cosx` necessarily. Functions are not always the same as constants.

`:.` with no value for `x`, this is an impossible proof.