Find all x such that sinx = cos5x (2 Viewers)

[Pimpcess.Snaz]

Question 7 (i) from the 1981 HSC paper.. i don't have the answers and the question is driving me crazy, help?


Domain: 0 < x < π

 

[undalay]

edit; nvmm too hardd

 

[me121]

Well, this will give you an idea of how many solutions and aprox where they are.

 

[Pimpcess.Snaz]

oooh alright, cause i thought maybe i was just stupid lol but yeah thanks guys and good idea graphing it, should've thought of that..

you reckon we're allowed to solve those type of questions by graphing? like without any working out (when its actually possible to see the solution)

 

[me121]

oooh alright, cause i thought maybe i was just stupid lol but yeah thanks guys and good idea graphing it, should've thought of that..

you reckon we're allowed to solve those type of questions by graphing? like without any working out (when its actually possible to see the solution)

Well, depending on how easy it takes to graph, I like to graph it just to check that I have the correct number of answers.

But the above example is difficult to graph and you can't really find the numerical answers just be graphing it.

But if you have the correct answers with no working you should get full marks. Its just that if you make a mistake along the way you will get zero. Hence you should show your working in case you make a silly mistake then you may still get some marks.

 

[Pimpcess.Snaz]

ah... okay thanks heaps

 

[webby234]

sinx = cos (pi/2-x)

So we have cos (2kpi + pi/2 - x) = cos5x

5x = 2kpi + pi/2 - x
6x = 2kpi + pi/2
x = kpi/3 + pi/12
x = pi/12, 5pi/12, 9pi/12

EDIT: I'm missing a couple of solutions, give me a minute

Ok other solutions are cos (2kpi - (pi/2 - x)) = cos5x
2kpi - pi/2 + x = 5x
4x = 2kpi - pi/2
x = kpi/2 - pi/8
x = 3pi/8, 7pi/8

 

[Pimpcess.Snaz]

So we have cos (2kpi + pi/2 - x) = cos5x

I don't understand this bit but your method looks pretty genius

 

[webby234]

Sorry, should have made that more clear...

It's the general solution: cosx = cos(2kpi +/- x)

so in this case we have cos(pi/2 - x) = cos (2kpi +/- (pi/2 - x))

 

[Pimpcess.Snaz]

ahh.. i see.. that's quite genius and here i was expanding cos5x. thanks webby

i can see my pyscho of a maths teacher putting that into a 3 unit trial...

 

[A High Way Man]

sif n0t know sin(x) = cos(pi/2 - x) f00lz

 

[Pimpcess.Snaz]

sif n0t know sin(x) = cos(pi/2 - x) f00lz

Translation: As if we didn't know sinx = cos(pi/2 - x) fools?

Umm.. we do know it -I just didn't think of using it in this question.