Find all the solutions of the Equation...

[mjlogan]

The question is

a) show that the equation

2sin²x – cosx = 1

can be written as

2cos²x + cosx – 1 =0

b) using your answer to part (a), find all the solutions of the equation

2sin²x – cosx = 1

in the interval 0 < x < 2ð, giving your answer in the terms of ð

the funny ð is supposed to be pi

my working so far to follow

[mjlogan]

a)

2sin²x – cosx = 1
2(1- cos²x)-cosx=1
2-2cos²x-cosx=1
2cos²x+ cosx-1=0

b)

(2cosx-1)(cosx+1)
2cosx=1 or cosx=-1
cosx=0.5 or cosx=-1

and now iam comfused… ive got no idea how iam supposed to go on from here

[sunburnt]

Look at the unit circle. ( cos(theta), sin(theta) )
http://www.math.lsa.umich.edu/~zacht...it_circle.html
When is the cos(theta) = 1/2?
When theta is 60 degrees, or pi/3, and also when theta is 5pi/3.

When is the cos(theta) = -1?
When theta is pi.

Of course, if you're using a calculator, you could use the arccos function, but that would be cheap :P and it probably wouldn't give you both answers for cos(theta) = .5

[mjlogan]

thank you... ive got my three answers from the diagram... but how is it possible to get the results without having that diagram.... for when i sit an exam... i will not be able to remember that whole diagram...my calculator does not give all 3 answers you are right it only gives 2....

thank you as well

[sunburnt]

We had to memorize the coordinates. :-)
There are three families. All you really need to know is the first one (pi/x), because the rest are simply sign changes by quadrant.

PI/4 FAMILY: multiples of 45 degrees
pi/4 (root2/2, root2/2) (+,+) 1st quadrant
3pi/4 (- root2/2, root2/2) (-,+) 2nd quadrant
5pi/4 (- root2/2, - root2/2) (-,-) 3rd quadrant
7pi/4 (root2/2, - root2/2) (+,-) 4th quadrant

PI/3 FAMILY multiples of 60 degrees
pi/3 (1/2, root3/2)
2pi/3
4pi/3
5pi/3

PI/6 FAMILY: multiples of 30 degrees
pi/6 (root3/2, 1/2) ....
5pi/6
7pi/6
11pi/6

I had trouble remembering what the multiples were, so I look at it like this:

PI/X
(X-1)PI/X
(X+1)PI/X
(2X-1)PI/X

hope this helps. Sorry, I hated memorizing them too. Maybe your teacher will let you use a sheet on the exam.

Note: when I say root2/2 and root 3/2 I mean sqr(2)/2 and sqr(3)/2, not sqr(2/2) or sqr(3/2).

[mjlogan]

okey dokey....cheers i will give it a go

[[Digital-X-Treme]]

Originally posted by mjlogan
thank you... ive got my three answers from the diagram... but how is it possible to get the results without having that diagram.... for when i sit an exam... i will not be able to remember that whole diagram...my calculator does not give all 3 answers you are right it only gives 2....

thank you as well

Learn the diagram! There are 4 quadrants on the diagram. All you have to do is remember the quadrants where Sin, Cos, Tan, and All functions are positive. The rest are then negative.

[IWright]

(2cosx-1)(cosx+1)
2cosx=1 or cosx=-1
cosx=0.5 or cosx=-1

I've been reading the replies for this post, and everything looks a bit complicated... surely if you want to find the x values you just need to do the following: (I'm assuming this is an AS/A level maths exam and u're using a scientific calculator):

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Find the inverse cos of both values to find x, in each case

cosx = 0.5 ==> x = 60degrees
cosx = -1 ==> x = 180degrees

draw the graph of y=cosx between 0 and 360degrees

mark the point at 60degrees where y = 0.5, draw a line if you need across the graph at y = 0.5, and you'll see theres another point on the graph where cosx = 0.5.

As the graph is symmetrical at x = 180degrees, and we know the first value of x is 60.

(180-60) = 120 ==> Therefore 180-120 = 60 degrees
hence 180+120 = 300degrees (or 360-60)

This shows that cos300 is the same value as cos60 (0.5)
So the first solutions are 60,300

Then Inverse cos of -1, you should know = 180degrees. On the cos graph there is only 1 point which reaches -1 between 0 and 360 degrees, so this is the only solution for that part.

All that is left is to convert 60,180,300 into pi*radians
180 = pi radians of course
60 = 180/3.. or pi/3 radians

300 = (60*5) = 5pi/3 radians
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I hope that helps, and at least for me it looks a little more friendly and like the expected answers from the examiners for As/A level exams