General Solutions

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August 2nd 2008, 12:02 AM
Misa-Campo

General Solutions

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

thanks
August 2nd 2008, 01:24 AM
mr fantastic

Quote:

Originally Posted by Misa-Campo

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

thanks

$\cos^2 x - \sin x \cos x = 0 \Rightarrow \cos x (\cos x - \sin x) = 0$ .

Case 1: $\cos x = 0$

$\Rightarrow x = \frac{(2n+1) \pi}{2}$ where n is an integer, that is, $x = \pm \frac{\pi}{2}, ~ \pm \frac{3\pi}{2}, ~ \pm \frac{5\pi}{2}, ~ ....$

Case 2: $\cos x - \sin x = 0 \Rightarrow \sin x = \cos x \Rightarrow \tan x = 1$

$\Rightarrow x = \frac{\pi}{4} + n \pi = \frac{(4n + 1) \pi }{4}$ where n is an integer, that is, $x = \frac{\pi}{4}, ~ \frac{5\pi}{4}, ~ \, ...., ~ - \frac{3 \pi}{4}, ~ \frac{-7 \pi}{4}, ~ ....$

It's the appearance of the n that makes the solution general.
August 2nd 2008, 01:44 AM
ticbol

Quote:

Originally Posted by Misa-Campo

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

thanks

General solutions? Let us find first the solutions.

(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.