sin theta/1+ cos theta=1/√3 find theta
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let theta =@
sin@/(1+cos@)=1/root3
root3sin@=1+cos@
root3sin@-cos@=1
2 (root3/2.sin@-1/2cos@)=1
2 (cospi/6.sin@-sinpj/6.cos@)=1
sin (@-pi/6)=1/2=sinpi/6
@-pi/6=n.pi+(-1)^n (pi/6)
@=n.pi+(-1)^n(pi/6)+pi/6
where n is integer numbers
sin@/(1+cos@)=1/root3
root3sin@=1+cos@
root3sin@-cos@=1
2 (root3/2.sin@-1/2cos@)=1
2 (cospi/6.sin@-sinpj/6.cos@)=1
sin (@-pi/6)=1/2=sinpi/6
@-pi/6=n.pi+(-1)^n (pi/6)
@=n.pi+(-1)^n(pi/6)+pi/6
where n is integer numbers
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