How do you find the general solution: #tanalpha - sin alpha = tanalphasinalpha# ?

1 Answer
Jan 11, 2018

#alpha = 2pin, pi + 2pin#

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 #pin#, because we're talking about #sin(2alpha)#, but that's false since clearly #alpha = pi# doesn't satisfy the equation.

Therefore,

#alpha = 2pin or pi/2 + 2pin#

However, the tangent function is not defined at #pi/2#, therefore, we must eliminate that solution.

Hopefully this helps!