How do you find the general solution: #tanalpha - sin alpha = tanalphasinalpha# ?
1 Answer
Explanation:
#sinalpha/cosalpha - sin alpha = sin alpha/cosalpha * sinalpha#
#(sin alpha - cosalphasinalpha)/cosalpha = sin^2alpha/cosalpha#
#sinalpha - cosalphasinalpha = sin^2alpha#
#0 = sin^2alpha - sin alpha + cosalphasinalpha#
#0 = sinalpha(sin alpha - 1 + cosalpha)#
So now we have two equations:
#sinalpha = 0 -> alpha = 2pin or pi + 2pin#
For the second:
#sinalpha - 1 + cosalpha = 0#
#sinalpha + cosalpha = 1#
#(sin alpha + cosalpha)^2 = 1^2#
#sin^2alpha + cos^2alpha + 2sinalphacosalpha = 1#
#1 + 2sinalphacosalpha = 1#
#sin(2alpha) = 0#
#2alpha = 0 or pi#
#alpha = 0 or pi/2#
You would think the period here is
Therefore,
#alpha = 2pin or pi/2 + 2pin#
However, the tangent function is not defined at
Hopefully this helps!