Complex number q (1 Viewer)

[abc123]

I've got no idea how to do this. Anyone know?

xcos(theta) + xsin2(theta) + 1= 0

 

[Bambul]

You have to get it all in terms of Sin(theta) or Cos(theta). I haven't done trig for about 1 1/2 years, so I'm a bit sketchy on this, but Sub Cos^2(theta) for 1 - Sin^2(theta) and then expand Sin(2theta). I forgot what Sin(2theta) expanded to. Then you can solve with the quadratic formula

 

[McLake]

theata = @

xcos@ + xsin2@ + 1= 0
x(1 - sin@) + xsin2@ + 1= 0

now use quadratic formula:

x = [-sin2@ +/- sqrt((sin2@) - 4*(1 - sin@)*1)]/2*(1 - sin@)
x = [-sin2@ +/- sqrt((4sin@cos@) - 4 - 4sin@)]/(2 - 2sin@)
x = [-sin2@ +/- sqrt((4sin@(cos@ - 1) - 4)]/(2 - 2sin@)

Hmm, I don't know, let me work on it ...

 

[OLDMAN]

Working on Bambul's and McLake's suggestion,

cos^2@x^2+2sin@cos@x+1=0 @not= pi/2,-pi/2 (meaningless otherwise)

x= (-2sin@cos@+-sqrt(4sin^2@cos^2@-4cos^2@)/2cos^2@)
= -tan@+-sqrt(4cos^2@(sin^2@-1))/2cos^2@)
= -tan@+- sqrt(-1) since sin^2-1=-1cos^2
= -tan@+-i

Another interesting method is by polynomial theory, watch:

quadratic has minimum at -b/2a=-tan@
minimum is sin^2@-2sin^2@+1=1-sin^2@>0, @ not Pi/2,-pi/2

Thus roots are complex and conjugate. Thus

a+ib+a-ib=2a=-2tan@ , therefore a =-tan@and
a^2+b^2=tan^2@+b^2=sec^2@ which gives b=+-1

This question is a good review of polynomial theory and trigonometric identities.

 

[spice girl]

Yet another way: completing the square

(cos^2@)(x^2) + (sin2@)x + 1 = 0

(cos^2@)(x^2) + (2sin@cos@)x + sin^2@ + cos^2@ = 0

(xcos@ + sin@)^2 + cos^2@ = 0

(xcos@ + sin@ + icos@)(xcos@ + sin@ - icos@) = 0

x = (-sin@ - icos@)/cos@, (-sin@ + icos@)/cos@

= -tan@ +- i