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

fortun
Jan 17, 2015

#6**+15 **

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

here is your equation

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

#1**+15 **

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

CPhill
Jan 17, 2015

#3**+5 **

I really like the question too. You were asking us to help you understand.

We really like such questions.

Thank you fortun and welcome to Web2 forum.

Melody
Jan 17, 2015

#5**+5 **

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?

fortun
Jan 17, 2015

#6**+15 **

Best Answer

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

here is your equation

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**+5 **

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

CPhill
Jan 17, 2015

#10**+5 **

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!!!!!!!!!

fortun
Jan 17, 2015

#11**+10 **

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.

Melody
Jan 17, 2015

#12**+5 **

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

fortun
Jan 17, 2015

#13**+5 **

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.

Melody
Jan 17, 2015