(E) : tan(x).tan(2x-π/3)=-1
or tan(a+b)=(tan(a)+tan(b))/(1-tan(a).tan(b))
donc 1-tan(a).tan(b)=(tan(a)+tan(b))/(tan(a+b))
donc tan(a).tan(b)=1-(tan(a)+tan(b))/(tan(a+b))
donc on déduit que (E) donne :
1-(tan(x)+tan(2x-π/3))/tan(3x-π/3)=-1
soit (tan(x)+tan(2x-π/3))/tan(3x-π/3)=2
donc tan(x)+tan(2x-π/3)=2tan(3x-π/3)
donc tan(2x-π/3)=tan(π/2+x)
donc x=-π/6+kπ ou x=11π/6+kπ
