A gay trig question which youll find flippin easy but i dont flippin get (1 Viewer)

[ram14]

I was just revisin trig functions and ths question:

0 < x < 360

Find:

sin x + (root)3 cos x = 0

???

 

[ram14]

dude i dont think this is that complicated...its just simple trig fucntions...its like question 17 from Fitzpatrick 2 unit...

 

[ianc]

It's easy now that you've changed the question!!!! (originally the cos was missing)

Hope this helps!

 

[jyu]

dude i dont think this is that complicated...its just simple trig fucntions...its like question 17 from Fitzpatrick 2 unit...

The edited version is much simpler than the original one that you posted.
Couldn't you do such a 'simple' trig equation?

:wave:

 

[ram14]

thanks for the help...no i couldnt do such a simple trig question coz i missed all of preliminary year doin british curriculum maths which doesnt have this topic in it...so im tryin to lear it all myself so i do decent in my hsc

 

[dr baby beanie]

Good luck, with ur determination and us helping you'll do fine

 

[Forbidden.]

It's easy now that you've changed the question!!!! (originally the cos was missing)

Hope this helps!

I see your are armed with MathType or equivalent ...

 

[pLuvia]

I was just revisin trig functions and ths question:

0 < x < 360

Find:

sin x + (root)3 cos x = 0

???

Alternatively you could use the auxilary method but this is faster I think

 

[ianc]

I see your are armed with MathType or equivalent ...

hey there

I just use equation editor in word and paste it into a gif....


okay youre getting a bit mixed up about the quadrant thing.

When you get the tanx=-sqrt(3), do not use your calculator. It will only confuse you with that -60.....


Without worrying about the negative yet, think back to that triangle of special angles - what value for tanx gives sqrt(3)?? (The answer is 60.)

Now in this particular question, they wanted 0 < x < 360, and also you want the value(s) for x where tanx is negative. This is where the quadrants come in.

Tan is negative in the second and fourth quadrants (Remember All Stations To Central?)

So to get your values for x, you simply do those operations using 60 (not -60!!!!)
Second quadrant: 180 - 60 = 120
Fourth quadrant: 360 - 60 = 300


I hope what I said made some sort of sense! Feel free to ask if youre still not sure. Remember, don't use your calculator!!

Have a good one!


edit: just reworded the explanation a little

 

[AlvinCY]

The way you guys have done it, probably a good idea to check if the solution will give cos x = 0, which I don't think it does, cos in the process you've divided both sides by a cos x... and you should be careful not to devide by 0.

I've done it using the auxiliary angles method:

I’m using a dot followed by a follow slash as a root sign (./ )

sin x + ./3 cos x = 0

Let sin x + ./3 cos x = A sin (x + y)
sin x + ./3 cos x = A sin x cos y + A cos x sin y
therefore, A cos y = 1, A sin y = ./3
cos y = 1/A, sin y = ./3 / A

If you draw the appropriate triangle with y being the angle, 1 being the adjacent side, ./3 being the opposite side and A being the hypotenuse; you find that A² = 1² + (./3)², giving A = 2, and solving cos y = 1/A = ½ (in first quadrant – note the way you drew the triangle, it’s in the first quadrant), y = pi/3

therefore sin x + ./3 cos x = 2 sin (x + pi/3) = 0
we need to solve sin (x + pi/3) = 0
x + pi/3 = k pi for some positive integer k (this is the general solution for sin)
so x = - pi/3 + k pi

Letting:
k = 1, x = (2/3)*pi = 120 degrees
k = 2, x = (4/3)*pi = 240 degrees
k = 3, x = (8/4)*pi which lies outside the domain of 0 < x < 360.


Therefore the only 2 solutions are 120 and 240.