We have here a quadratic equation in #cosx#
#2cos^2x+3cosx-3=0#
Hence to get #cosx# we can use quadratic formula, where #a=2#, #b=3# and #c=-3# and as solution of quadratic formula is
#(-b+-sqrt(b^2-4ac))/(2a)#
Hence #cosx=(-3+-sqrt(3^2-4xx2xx(-3)))/4=(-3+-sqrt33)/4#
i.e. #(-3-sqrt33)/4# and #(-3+sqrt33)/4#
As #|(-3-sqrt33)/4|=(3+5.745)/4>1# and hence is not possible and
we can only have #cosx=(-3+5.745)/4=2.745/4=0.6852#
and #cosx=cos46.75^@# from scientific calculator or tables.
But as #cos(-x)=cosx# we can have #x=46.75^@# or #x=-46.75^@#
Further, every trigonometric ratio has a cycle of #360^@#
Hence #x=nxx360^@+-46.75^@#, where #n# is an integer.
