trig general solutions (1 Viewer)

[NexusRich]

How do I find the general solution if the term is negative, like in case, it is -1/2. Do I find the cos inverse of -1/2 - which is in 2nd quadrant ? and use the formula like usual ? It didnt work for some reason

 

[cossine]

Do I find the cos inverse of -1/2 - which is in 2nd quadrant ?

Yes do exactly that.


The process is the same.

=> cos^-1(-1/2) = 120 degrees = 2pi/3.

Remember the range of cos^-1(x) is [0 , pi] and domain [-1, 1]


So the general solution for

cos(x) = -1/2

is

x = 2*n*pi +/- 2pi/3 eq(1) (using theorem 1)


However we got cos(2x) = -1/2

So we replace x with 2x from eq(1)

=> 2x = 2*n*pi +/- 2pi/3

=> x = n*pi +/- pi/3

Therefore the answer is D
___________________________________


If the equation was cos(ax+b) = d

Then similarly we would just replace x with ax+b that is used in theorem 1

So the general solution will be given by

ax+b = 2*n*pi +/- cos^-1(d)


____________________________________
Theorem 1:

the general soultion for cos x = d

is given by

x = 2*n*pi +/- cos^-1(d)

 

[CaptainAmerica5050]

does anyone know if any schools has conducted their trials already?

 

[CM_Tutor]

The first thing to note with an MCQ like this one is that you don't actually need to find a general form.

The second is that there can be more than one general form.

Now, the equation here is . This has a related angle (acute) given by .

The solutions are in the 2nd and third quadrants the cosine function is negative. So, we have:

​

We can now choose the correct answer by looking at which option generates the same set of solutions:

(A)

(B)

(C)

(D)

And the answer is clearly (D).

We could have short cut this by looking at our solution set, which includes only multiples of that appear in all four quadrants and then recognise that (A) and (B) will give solutions only in quadrants 1 and 2, and (C) will give multiples of .


Having said all that, we can also find the general formula, in which case your key mistake is not realising that the related angle must be acute whether the function is positive or negative, and you adjust for quadrants from there.

In other words, we seek with a related angle , but in quadrants 2 and 3 (as cos is negative). In other words:

​

This is why there are the options with in them...

Anyway, if is ODD, then I can set where is any integer:

​

Now, since takes the value of every integer, and so therefore does , I can combine these and generate the same solutions, though in a different order, as , which is the answer given in the question. However, if an exam question sought a general formula, I think that is fine, so long as you clearly state the requirement that is ODD. In the alternative, I also think that for all integers is a reasonable answer.

 

[CM_Tutor]

The third thing to note is that CM_Tutor sometimes goes the long way when a shorter option is available, like the much more direct and totally valid approach from @cossine that finds the given answer quickly!