I have no idea about general solutions since i have never seen them, What are all the general solutions that i must memorize to answer all questions?
and how do i apply these general solutions in this question:
Find all values of x for which (cosx)^2=sinxcosx
General solutions? Let us find first the solutions.
Originally Posted by Misa-Campo
(cosX)^2 = sinXcosX
(cosX)^2 -sinXcosX = 0
cosX(cosX -sinX) = 0
cosX = 0
X = arccos(0) = pi/2, and/or 3pi/2 -------**
cosX -sinX = 0
cosX = sinX
Divide both sides by cosX,
1 = tanX ------------positive tan value, so X is in the 1st and 3rd quadrants.
X = arctan(1)
X = pi/4, and/or (pi +pi/4) = 5pi/4 .................***
Those solutions, X = pi/2, 3pi/2, pi/4, 5pi/4 are slutions in one revolution only, or from 0 to 2pi only.
Maybe by general solutions it means for all the possible values of X.
If so, then,
a) when cosX = 0,
the period of the basic cosine curve y = cosX is 2pi. So the X = pi/2 and 3pi/2 repeats every 2pi. Hence, the general solutions when cosX = 0 are
X = pi/2 +n(2pi)
and, X = 3pi/2 +n(2pi)
where n is any integer.
b) when cosX -sinX = 0, or when tanX = 1,
the period od the basic y = tanX curve is pi only.
Hence the general solutions when tanX = 1 are
X = pi/4 +n(pi)
and X = 5pi/4 +n(pi)
But when n=1, pi/4 +1(pi) = 5pi/4 .....which is 5pi/4 +0(pi).
And 5pi/4 +1(pi) = 9pi/4 .....which is pi/4 +2(pi).
Meaning they overlap or are the same angles.
So, for tanX = 1, the general solution is X = pi/4 +n(pi) only.