Question 874260
{{{tan(2theta-34^o)=sqrt(3)}}} has infinite solutions, unless the interval is restricted.
{{{tan(60^o)=sqrt(3)}}} ,
so {{{2theta-34^o=60^o}}}-->{{{2theta=60^o+34^o}}}-->{{{2theta=94^o}}}-->{{{theta=94^o/2}}}-->{{{theta=47^o}}} is one answer.
However, tangent has a period of {{{180^o}}} , so many other answers are possible.
In general,
{{{tan(60^o+k*180^0)=sqrt(3)}}} for any {{{k}}} integer.
So all solutions can be found from
{{{2theta-34^o=60^o+k*180^0}}}-->{{{2theta=60^o+k*180^0+34^o}}}-->{{{2theta=94^o+k*180^0}}}-->{{{theta=(94^o+k*180^0)/2}}}-->{{{theta=94^o/2+k*180^0)/2}}}-->{{{theta=47^o+k*90^0}}}
In the interval {{{"[ 0 ,"}}}{{{360^o}}}{{{")"}}} the solution is
for {{{k=0}}} , {{{theta=47^o}}} ,
for {{{k=1}}} , {{{theta=37^o}}} ,
for {{{k=2}}} , {{{theta=227^o}}} ,
for {{{k=3}}} , {{{theta=317^o}}} .