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

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.