Wie löst man #cos2x = sinx # im Intervall #0 <= x <= 2pi #?

If #cos(2x) = sin(x)#
dann
#1-2sin^2(x) = sin(x)#

#2sin^2(x) +sin(x) -1 =0#

Substitution #k=sin(x)#
#2k^2+k-1 = 0#

#(2k-1)(k+1) = 0#

#sin(x) = 1/2# or #sin(x) =-1#

If #sin(x) = 1/2# (Für #0<=x<=2pi#)
#x=pi/6 =30^o# or #x= (5pi)/6 = 150^o#

If #sin(x) = -1# (Für #0<=x<=2pi#)
#x=(3pi)/2 = 270^o#

So #x epsilon {pi/6, (5pi)/6, (3pi)/2}#
(oder ihr Äquivalent in Grad)