7 replies to this topic

[Adam]

root3cosx + sinx = root2

therefore 2cos(x - /6) = root2

cos(x - /6) = root2/2 = 1/root2

(x - /6) = /4 or 7/4

My question is, where does the 7/4 bit come from? I understand where the /4 comes from, but not 7/4. Can somebody please explain?

Thank-you.

[Mr H]

From the CAST rule.

45 is the acute angle.

Use the CAST diagram to find the second solution for positive cos.

i.e. in the 4th quadrant.

So 2nd solution (in degrees for easier illustration) is 360 - 45 = 315 .

Which is 7 x 45 , in other words 7 /4.

[Adam]

Hmm.. I think I understand. But what about this one?

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?

(the final answers in this question are 0, 4/3 and 2 :s)

[ad absurdum]

Hmm.. I think I understand. But what about this one?

cosx - root3sinx = 1

k = 2
alpha = 300

So 2cos(x - 300) = 1

-> cos(x - 300) = 0.5
-> x - 300 = 60, 300?

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Say your looking for values of x that satisfy this equation with the range:


Now we have:




Now if we took the normal values for the inverse cosine of 0.5 we would get the angle of 60^o, and we would say that within the range the values are 60^o and 300^o
However, we are adding 300^o onto the angle, and if we add 300^o onto 300^o then we get a value outwith our range.
However, adding 300^o onto 60^o gives us a value still within the range so it's fine to use this
There is another value for x though, and if you think of this as coming from the cosine graph before zero (because it repeats, just as there is a repeat of the graph after 360 there is one before zero). So if we took -60 (i.e. the equivalent of the quadrant four angle in the graph before zero) as one of our values for x, added 300 onto this and got 240 this would give us a value still within our range. Also, if we took the quadrant one angle from our graph before zero, which is -300 (i.e. -360 + 60), and add 300 onto this we get zero which is still in our range.
So from this we get the results:
x = 0, 240, 360

That's how I do it anyway, not sure about how good a method it is

[Mr H]

Beat me to it ad adsurdum!

Too slow at typing!!

Following from your last line Betinho;

x = 360 , 600

As ad adsurdum states, 600 is outwith the usual 0 < x < 360 domain.

The period/wavelength is 360 as there is only a single 'x' therefore every 360 there will be another solution.

The 360 solution above is OK, there will also be a solution 360 before it, i.e. 0 , and also 360 before 600 i.e. 240 .

So the solutions are (in degrees) 0 , 240 , 360

Or 0, 4 /3, 2 in radians.

[Mr H]

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Your method is fine ad adsurdum!

[Adam]

Thanks guys!

It makes sense now. It was beginning to annoy me - I've been staring at that question for days.

[ad absurdum]

I hate these type, and I've never been told the proper way to solve them so I do it with a bit of common sense.

Your method is fine ad adsurdum!

Ah good to hear, I was worried that there was a quicker way I was failing to notice