Trig Equation

I've solved the attached equation and got 5pi/20 and 13pi/20, but was informed there are three more solutions. From obtaining 5x - pi/4= pi + n2pi where n is an integer I just followed on algebraically until I got my two solutions. How do I obtain the missing ones?

Original post by samjohnny

I've solved the attached equation and got 5pi/20 and 13pi/20, but was informed there are three more solutions. From obtaining 5x - pi/4= pi + n2pi where n is an integer I just followed on algebraically until I got my two solutions. How do I obtain the missing ones?

Okay, so let's say I had

\sin x = 5

and I had to solve it in the range

0 < x \leq 360

(degrees), I would solve it normally, by using the sine graph or the CAST diagram.

However, if I had

\sin (x+20) = 5

and I had to solve it in the range

0 < x \leq 360

(degrees), because I no longer am dealing with x - instead I'm dealing with x+20 - my range must change. It will now become:

20 < x+20 \leq 380 (degrees)

Hence, I have to extend my sine graph (or continue to use the CAST diagram) to find other solutions in that range. Do you get what I'm trying to say? So, what would your range become?

You also might want to check the two answers you have - I don't quite understand what you've done.

(edited 9 years ago)

Reply 2

brianeverit

9

Original post by samjohnny

I've solved the attached equation and got 5pi/20 and 13pi/20, but was informed there are three more solutions. From obtaining 5x - pi/4= pi + n2pi where n is an integer I just followed on algebraically until I got my two solutions. How do I obtain the missing ones?

0\leq\alpha\leq\pi \implies 0\leq5\alpha\leq5\pi \implies 0\leq 5\alpha-\frac{\pi}{4}\leq\frac{19\pi}{4}

so we must consider

5\alpha-\frac{\pi}{4}=0,\ \pi,\ 2\pi,\ 3\pi,\ 4\pi

giving 5 solutions

Reply 3

samjohnny

OP

Ah right, I get it now. Thanks a lot.

I'm also having a hard time on the one after it. It's attached. Not sure where to start or how to follow through.

(edited 9 years ago)

trig question.JPG20.6KB

Reply 4

davros

16

Original post by samjohnny

Ah right, I get it now. Thanks a lot.

I'm also having a hard time on the one after it. It's attached. Not sure where to start or how to follow through.

You want to use the hint they've provided to combine a couple of the cosine terms. Have a play with it and see if you can work out which two can be usefully combined.

Reply 5

ztibor

10

Original post by samjohnny

I've solved the attached equation and got 5pi/20 and 13pi/20, but was informed there are three more solutions. From obtaining 5x - pi/4= pi + n2pi where n is an integer I just followed on algebraically until I got my two solutions. How do I obtain the missing ones?

Firstly
You should to consider that the sine function is periodic and to solve
the equation according to this generally

e.g. 1. When

\displaystyle \sin \left (5\alpha -\frac{\pi}{4}\right )=0

then

\displaystyle 5\alpha -\frac{\pi}{4}=k\cdot 2\pi

or

\displaystyle 5\alpha -\frac{\pi}{4}=\frac{3\pi}{4}+k\cdot 2\pi

That is
First solution sequence

\displaystyle \alpha = \frac{\pi}{20}+k\cdot \frac{2\pi}{5}

and the second solution sequence

\displaystyle \alpha = \frac{\pi}{5}+k\cdot \frac{2\pi}{5}

Secondly.
From general solutions you have to choose those whiches are in the
given range for

k=0, \pm 1, \pm 2 ...

(edited 9 years ago)

Trig Equation

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