solve the following problem (√3+1)*(cos(x/2))^2+((√3-1)sinx))/2-1=0 (-π<x<π)
right answer
tan(x/2)=s (-π/2< x/2 < π/2)
1+(tan(x/2))^2=1/(cos(x/2))^2) , cos(x/2))^2=1/(1+s^2)
sinx=2s/(1+s^2)
(√3+1)/(1+s^2)+((√3-1)/2)*(2s/(1+s^2))-1=0
∴ (√3+1)+(√3-1)s-(1+s^2)=0
∴ s^2-(√3-1)s-√3=0
∴ (s+1)*(s-√3)=0
tan(x/2)=-1, √3, (x/2)=-π/4, π/3
∴ x=-π/2, 2π/3
what i have done
(√3+1)*(cos(x/2))^2+((√3-1)sinx))/2-1=0 (-π<x<π)
(√3+1)*(cosx+1)/2+((√3-1)sinx))/2-1=0
(√3+1)cosx+(√3-1)sinx-2+√3+1=0
(√3+1)cosx+(√3-1)sinx=1-√3
(4+2√3)(cosx)^2+(4-2√3)(sinx)^2+2(√3+1)(√3-1)cosx*sinx=4-2√3
4((cosx)^2+(sinx)^2)+2√3((cosx)^2-(sinx)^2)+2sin(2x)=4-2√3
2√3((cosx)^2-(sinx)^2)+2sin(2x)+2√3=0
2√3(cos(2x))+2sin(2x)+2√3=0
cos(2x)+(1/√3)*sin(2x)+1=0
tanx=t (-π<x<π)
cos(2x)=(1-t^2)/(1+t^2), sinx=2t/(1+t^2)
cos(2x)+(1/√3)*sin(2x)+1=0
(1-t^2)/(1+t^2)+(1/√3)*(2t/(1+t^2))+1=0
(1+t^2)>0
(1-t^2)+(1/√3)*2t+(1+t^2)=0
t=-√3
∴ tanx=-√3
∴ x=-π/3, 2π/3
Why did the wrong answer come out?:sad:
right answer
tan(x/2)=s (-π/2< x/2 < π/2)
1+(tan(x/2))^2=1/(cos(x/2))^2) , cos(x/2))^2=1/(1+s^2)
sinx=2s/(1+s^2)
(√3+1)/(1+s^2)+((√3-1)/2)*(2s/(1+s^2))-1=0
∴ (√3+1)+(√3-1)s-(1+s^2)=0
∴ s^2-(√3-1)s-√3=0
∴ (s+1)*(s-√3)=0
tan(x/2)=-1, √3, (x/2)=-π/4, π/3
∴ x=-π/2, 2π/3
what i have done
(√3+1)*(cos(x/2))^2+((√3-1)sinx))/2-1=0 (-π<x<π)
(√3+1)*(cosx+1)/2+((√3-1)sinx))/2-1=0
(√3+1)cosx+(√3-1)sinx-2+√3+1=0
(√3+1)cosx+(√3-1)sinx=1-√3
(4+2√3)(cosx)^2+(4-2√3)(sinx)^2+2(√3+1)(√3-1)cosx*sinx=4-2√3
4((cosx)^2+(sinx)^2)+2√3((cosx)^2-(sinx)^2)+2sin(2x)=4-2√3
2√3((cosx)^2-(sinx)^2)+2sin(2x)+2√3=0
2√3(cos(2x))+2sin(2x)+2√3=0
cos(2x)+(1/√3)*sin(2x)+1=0
tanx=t (-π<x<π)
cos(2x)=(1-t^2)/(1+t^2), sinx=2t/(1+t^2)
cos(2x)+(1/√3)*sin(2x)+1=0
(1-t^2)/(1+t^2)+(1/√3)*(2t/(1+t^2))+1=0
(1+t^2)>0
(1-t^2)+(1/√3)*2t+(1+t^2)=0
t=-√3
∴ tanx=-√3
∴ x=-π/3, 2π/3
Why did the wrong answer come out?:sad: