How do you solve #\frac { 2\sin \frac { 5} { 6} \pi + \cos x } { \tan \frac { \pi } { 2} + \cos \frac { 2\pi } { 3} } = 0#?

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Answer 1 · 2018-05-10T17:09:05

#x=(2k-1)pi, kinZZ#

We have,

#(2sin((5pi)/6)+cosx)/(tan(pi/2)+cos((2pi)/3))=0#

#=>2sin((5pi)/6)+cosx=0#

#=>2sin(pi-pi/6)+cosx=0#

#=>2sin(pi/6)+cosx=0#

#=>2xx1/2+cosx=0#

#=>1+cosx=0#

#=>cosx=-1#

#=>x=(2k-1)pi, kinZZ#

Note:

#(i)# If, # 0 <= x < 2pi, then,x=pi#

#(ii)tan(pi/2)# is undefined and right hand side is given zero, then denominator #=tan(pi/2)+cos((2pi)/3)# is meaningless .!?