by using polar form and de moivres theorem, (cosx + jsinx)*3/ cos2x - jsin2x?

Answer 1

acc to de moiver's theorem
(cosx+jsinx)^k=coskx+jsinkx
therfore
cos2x-jsin2x=cos(-2x)+jsin(-2x)
=(cosx+jsinx)^(-2)
therfore
p=3*(cosx+jsinx)/(cosx+jsinx)^-2
=3(cosx+jsinx)^2
=3cos2x+jsin2x
thanks & good luck

Answer 2

de Moivre's identity asserts that,

(cos z + i sin z)^n = cos nz + i sin nz , apparent from Euler's relation and absolutely true by induction for integer n.
then,
3(cos z + i sin z) / (cos 2z - i sin 2z)
= 3(cos z + i sin z) / (cos z + i sin z)^(-2)
= 3 (cos z + i sin z)^[1-(-2)]
= 3 (cos 3z + i sin 3z)

I have treated 3 as a coefficient and not an exponent.
i has been used instead of j, and z instead of x

Answer 3

(cosx + jsinx)*3/ (cos2x - jsin2x) NOTE: I assume you meant this
= 3e^(jx)/e^(-2jx)
= 3e^(jx) * e^(2jx)
= 3e^(3jx)
= 3*(cos3x + jsin3x)