Question #2fa4f

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Answer 1 · 2017-12-21T17:02:57

#"the Solution Set="{2/9pi,5/9pi,8/9pi,11/9pi,14/9pi,17/9pi}#.

#sqrt3cot3x+1=0#.

#:. cot3x=-1/sqrt3#.

#:. tan3x=-sqrt3=tan(-pi/3)#.

Since, #tantheta=tanalpha rArr theta=kpi+alpha, k in ZZ#, we have,

# 3x=kpi-pi/3, k in ZZ, or, x=(3k-1)pi/9, k in ZZ#.

Now, #0 le x le 2pi rArr 0 le (3k-1)pi/9 le 2pi#,

#rArr 0 le (3k-1)pi le 18pi#,

#rArr 0 le (3k-1) le 18#,

#rArr 1 le 3k le 19#,

#rArr 1/3 le k le 19/3=6 1/3#.

As #k in ZZ, k=1,2,3,...,6#.

Accordingly, #"the Solution Set="{2/9pi,5/9pi,8/9pi,11/9pi,14/9pi,17/9pi}#.