help

Try this
1. tan(2a)+2cos(a)=0
sin(2a)/cos(2a)+2cos(a)=0
using half angle formulas
2sin(a)cos(a)/[cos^2(a)-sin^2(a)]+2cos(a)=0
2sin(a)cos(a)+2cos(a)[cos^2(a)-sin^2(a)]=0
cos(a)(cos^(a)+sin(a)-sin^2(a))
from sin^(a)+cos^2(a)=1
cos(a)(2sin^2(a)-sin(a)-1)=0
Factoring
cos(a)(2sin(a)-1)(sin(a)+1)=0
roots are
cos(a)=0
sin(a)=1/2
sin(a)=-1

2) deMoivre Theorem
z^n=r^n[cos(n*theta)+isin(n*theta)]
Put -8+8*3^(1/2)i) into form deMoivre to zero
z=-8+8*3^(1/2)i=16(-1/2+sqrt(3)/2i)
z=16(cos(240)+sin(240))
z^(1/4)=16(1/4)(cos(1/4*240)+isin(1/4*240))
z^(1/4)=2(cos(60)+isin(60))
z^(1/4)=2(1/2+sqrt(3)/2)i)=1+sqrt(3)i

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Thank you.
And how could you solve this one:
sin(2a) + cos(2a) = 0

Use the half angle formulas and factor
sin(2a)=2cos(a)sin(a)
cos(2a)=cos^2(a)-sin^2(a)
so
sin(2a)+cos(2a)=0
becomes
2sin(a)cos(a)+cos^2(a)-sin^2(a)=0
let x=sin(a) and y=cos(a)
this becomes
2xy+y^2-x^2=0
rearranging
y^2+2xy-x^2=0
factor this expression find solutions for x and y
and solve those for a=arcos(y) and a=arcsin(x)

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