Question #071e3

2 Answers
NickTheTurtle
Apr 8, 2017

#x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ#

Explanation:

If #ab=0#, then either #a=0# or #b=0# or #a=b=0#.

Looking at the equation #(sin(x)+3)(2sin(x)+1)=0#, it can be said that the solutions occur when #sin(x)+3=0# or #2sin(x)+1=0#.

Consider #sin(x)+3=0# first. The range of #sin(x)# is #-1≤sin(x)≤1# for all #x in RR#. There are no solutions here.

Consider #2sin(x)+1=0#. This is equivalent to #sin(x)=-1/2#. Solving for #x# in the range #[0,2pi)#, we get #(7pi)/6# and #(11pi)/6#.

Since #sin(x)# is periodic with period #2pi#, the solution is #x=(7pi)/6+2kpi vv x=(11pi)/6+2kpi,k in ZZ#.

Alan P.
Apr 8, 2017

#x=(7pi)/6+2kpi" or "x=(11pi)/6+2kpi#
#color(white)("XXXXXXXXXXXXXXXXXXXXX")AAk in ZZ#

Explanation:

If #(sin(x)+3)(2sin(x)+1)=0#
#{: ("either ",sin(x)+3=0," or ",2sin(x)+1=0), (,"impossible",,rarr sin(x)=-1/2), (,color(white)("X")"since "sin(x) in[-1,+1],,x=(7pi)/6" or "(11pi)/6), (,,,color(white)("X")"assuming "x in [0,2pi)) :}#

Detail:
#color(white)("XXX")sin(x)=-1/2#
implies a reference angle of #pi/6# (one of the standard angles)
in either Quadrant III or IV

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