How do you solve #3sinx+2=tan75#?

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Answer 1 · 2016-07-09T02:27:14

#:.x=npi+(-1)^nalpha#
where #sin^-1(1/sqrt3)=alpha#

#3sinx+2=tan75=(tan30+tan45)/(1-tan30tan45)#

#=>3sinx+2=(1/sqrt3+1)/(1-1/sqrt3xx1)#

#=>3sinx+2=(sqrt3+1)/(sqrt3-1)#

#=>3sinx+2=(sqrt3+1)^2/((sqrt3-1)(sqrt3+1)#

#=>3sinx+2=(4+2sqrt3)/(3-1)#

#=>3sinx+2=2+sqrt3#

#=>3sinx=sqrt3#

#=>sinx=sqrt3/3=1/sqrt3==sinalpha#

where #sin^-1(1/sqrt3)=alpha#

#:.x=npi+(-1)^nalpha#

#"where "n=+-1,+-2,+-3....#