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I have solved the equation 2cosΘsinΘ=cosΘ where [0,2π) and got the solutions (Π/2,Π/6) but my math textbook has p

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I typed the question wrongly earlier.I have solved the equation

2cosΘsinΘ=cosΘ where [0,2π)

and got the solutions (Π/2,Π/6) but my math textbook has provided two additional value to my solution

(5π/6,3π/2)

PLEEAASSEEE HEEELLLPPP!!!!!!!!!!!!

UPDATED

Jan 17, 2015

#6
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No you don't have to memorise anything.  Well not much anyway. lol

2cosΘsinΘ=cosΘ where [0,2π)

take it all to one side

2cosΘsinΘ-cosΘ=0

factorise

cosΘ(2sinΘ-1)=0

now  this will only be true if one of the factors =0

ie

cosΘ=0           $$\theta = \pi/2 \;\;or\;\; 3\pi/2$$

or

2sinΘ-1=0           $$sin\theta=1/2$$          $$\theta = \pi/6\;\;or\;\;5\pi/6$$

so

$$\theta = \pi/2 \;\;or\;\; 3\pi/2$$    $$\;\;or\;\;\pi/6\;\;or\;\;5\pi/6$$

Now if you are still having problems please try and tell us what you do not understand.

Jan 17, 2015

#1
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2cosΘsinΘ=cosΘ     subtract cosΘ from each side

2cosΘsinΘ - cosΘ  = 0      factor

cosΘ (2sinΘ - 1) = 0

And setting each factor to 0, we have

cosΘ = 0  and this happens at pi/2 and 3pi/2

And

sinΘ = 1/2    and this happens at pi/6 and 5pi/6

It looks like you just included the first quadrant values in your solution......

Here's a graph of both sides......https://www.desmos.com/calculator/crdklgepre

Jan 17, 2015
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Jan 17, 2015
#3
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We really like such questions.

Thank you  fortun and welcome to Web2 forum.

Jan 17, 2015
#4
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Thanx.....!!!

Jan 17, 2015
#5
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Thanks @CPhill.Appreciate your quick responsesThanks for the welcome@Melody.Ok so could you dumb it down a little for me?Are these graph values constant so I just have to memorize the graph or they have to be worked out every single time?

Jan 17, 2015
#6
+95177
+15

No you don't have to memorise anything.  Well not much anyway. lol

2cosΘsinΘ=cosΘ where [0,2π)

take it all to one side

2cosΘsinΘ-cosΘ=0

factorise

cosΘ(2sinΘ-1)=0

now  this will only be true if one of the factors =0

ie

cosΘ=0           $$\theta = \pi/2 \;\;or\;\; 3\pi/2$$

or

2sinΘ-1=0           $$sin\theta=1/2$$          $$\theta = \pi/6\;\;or\;\;5\pi/6$$

so

$$\theta = \pi/2 \;\;or\;\; 3\pi/2$$    $$\;\;or\;\;\pi/6\;\;or\;\;5\pi/6$$

Now if you are still having problems please try and tell us what you do not understand.

Melody Jan 17, 2015
#7
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The values generated by these solutions are ones associated with the pretty common "reference" angles  in the first quadrant [0, pi/6, pi/4, pi/3 and pi/2]....you should commit these to memory.......and then, be able to apply them in any of the other 3 quadrants....this is a VERY important trig topic....!!!

This might help.......http://www.coolmath.com/reference/circles-trigonometry.html

Jan 17, 2015
#8
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I do not memorise ANY of them.  I just understand them!

Jan 17, 2015
#9
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Here's something else that might help....the unit circle....

Jan 17, 2015
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Wow you guys are good.Thanks sooooo much!!!!No seriously,thanks so much.Av been stumped by this for quite a while n every other math site assumes i know trig already..lol.Thanks guys!!!!!!!!!

Jan 17, 2015
#11
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Wow Chris - Poor Fortun will think he is supposed to memorise all that!!

The unit circle is very useful but mostly just for

0, pi/2, pi, and 3pi/2 radians and multiples of these of course.

Are you really up for a full scale lesson here Fortun.

If you get me and probably Chris reved up you may not be able to stop us lol.

Or have you learned enough for the moment.   You can ask for more help whenever you want it either way.

Jan 17, 2015
#12
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Actually it seems the unit circle is really useful for this question and others I had...yes there were quite a lot more.Thanks guys

Jan 17, 2015
#13
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Please take a look at this unit of work.

http://www.mathsisfun.com/geometry/unit-circle.html

I do not like their explanation of tan much but their explanation of sine and cosine looks really good.

Try and understand the interactive pics properly.

Basically if you draw a unit circle (a circle of radius 1 unit) centred at (0,0) then

$$\\sin\theta = y \;value\\ cos\theta = x \;value\\ tan\theta= \frac{y}{x}=\frac{sin\theta}{cos\theta}$$

The (x,y) value that I am referring to is where the arm of the angle cuts the diametre of the unit circle.

The diagrams in the unit are trying to show you why this is so.

The more you understand the less you have to memorise and the easier it is to move on to the next step.

Jan 17, 2015