Trigonometric equation

[Kalen12]

I stumbled upon a problem which I'm not sure how to solve. It goes like this:

How many solutions are there to the equation sin2x-2sinx+cosx-1=0 on the interval (0,π)?

Here as far as I have been able to go:

sin2x-2sinx+cosx-1=0
2sinxcosx-2sinx+cosx-1=0
2sinx(cosx-1)+cosx-1=0
(2sinx+1)(cosx-1)=0

2sinx+1=0
2sinx=-1
sinx=-1/2
x₁ =7π/6 +2kπ
x₂ = 11π/6 +2kπ

So neither of these should be a solution, right? Since they belong to the III and IV quadrant and the interval is saying that it is in the I and II quadrant since from 0 to π is from 0 degrees to 180 degrees, correct me if I'm wrong please.

cosx-1=0
cosx=1

x=0+2kπ

Now this one is the most confusing for me since I don't know where this really belongs and does it fit in this interval of (0,π)?

 

[Deleted member 4993]

I stumbled upon a problem which I'm not sure how to solve. It goes like this:

How many solutions are there to the equation sin2x-2sinx+cosx-1=0 on the interval (0,π)?

Here as far as I have been able to go:

sin2x-2sinx+cosx-1=0
2sinxcosx-2sinx+cosx-1=0
2sinx(cosx-1)+cosx-1=0
(2sinx+1)(cosx-1)=0

2sinx+1=0
2sinx=-1
sinx=-1/2
x₁ =7π/6 +2kπ
x₂ = 11π/6 +2kπ

So neither of these should be a solution, right? Since they belong to the III and IV quadrant and the interval is saying that it is in the I and II quadrant since from 0 to π is from 0 degrees to 180 degrees, correct me if I'm wrong please.

cosx-1=0
cosx=1

x=0+2kπ

Now this one is the most confusing for me since I don't know where this really belongs and does it fit in this interval of (0,π)?

k = 0 → x = 0

 

[Kalen12]

k = 0 → x = 0

So what do you mean by that?
If x=0 then k=0 doesnt seem logical to me
And do any of the solution belong to that interval?

 

[pka]

I stumbled upon a problem which I'm not sure how to solve. It goes like this:
How many solutions are there to the equation sin2x-2sinx+cosx-1=0 on the interval (0,π)?
Here as far as I have been able to go:
sin2x-2sinx+cosx-1=0
2sinxcosx-2sinx+cosx-1=0
2sinx(cosx-1)+cosx-1=0
(2sinx+1)(cosx-1)=0
2sinx+1=0
2sinx=-1
sinx=-1/2
x₁ =7π/6 +2kπ
x₂ = 11π/6 +2kπ

We are told that \(\displaystyle 0\le x\le \pi\). Therefore, \(\displaystyle x\) is in the first or second quadrant.
The sine function is positive there. So there is no solution for \(\displaystyle 2\sin(x)=-1\).
There is one solution for \(\displaystyle \cos(x)=1\), what is it?

 

[Harry_the_cat]

You are correct in saying that there are no solutions to sin x = -1/2 in (0, pi) because in the first and second quadrant, sin x >0.

If cos x = 1 then x= 0 + 2k(pi) ie 0, 2(pi), 4(pi) … etc. Also -2(pi), -4(pi) etc. None of these are in (0, pi).

(Note that when Subho is saying "if k=0, then x=0" , he is finding one of the infinite solutions of cos x = 1. If k=1 then x = 2(pi) etc.)

So there are no solutions to the equation on the interval (0, pi).

 

[Harry_the_cat]

To PKA:
The round brackets, as opposed to square brackets, around (0, pi) means 0<x<pi.
That is, 0 and pi are not included. Using a less than or equal to sign is incorrect. And so there are no solutions in the given interval.

 

[pka]

To PKA:
The round brackets, as opposed to square brackets, around (0, pi) means 0<x<pi.
That is, 0 and pi are not included. Using a less than or equal to sign is incorrect. And so there are no solutions in the given interval.

I suspect that was a mistake in the post.

 

[Kalen12]

Well thanks a lot for the help. This is a lot clearer to me.

You are correct in saying that there are no solutions to sin x = -1/2 in (0, pi) because in the first and second quadrant, sin x >0.

If cos x = 1 then x= 0 + 2k(pi) ie 0, 2(pi), 4(pi) … etc. Also -2(pi), -4(pi) etc. None of these are in (0, pi).

(Note that when Subho is saying "if k=0, then x=0" , he is finding one of the infinite solutions of cos x = 1. If k=1 then x = 2(pi) etc.)

So there are no solutions to the equation on the interval (0, pi).

I got one question, if cos(x)=1 and our solution interval is 0≤ x≤ 4pi would then cos(x) be the solution since x=0+2kpi

 

[Harry_the_cat]

x = 0 + 2k(pi) is the general solution to the equation cos x = 1.
If k=0 then x = 0
If k=1 then x = 2(pi)
If k=2 then x = 4(pi)
These are the 3 solutions in the interval [0, 4(pi)]

 

[Steven G]

I too suspect that (0,pi) is not correct. I bet the problem stated that it included at least one of the end points.
In solving sin x = -1/2 on (0,pi) you really did not do the work which you did. As pointed out, if sin x <0, the x is in quad III or IV and NO angles in those quadrants are in (0,pi).

 

[Steven G]

Well thanks a lot for the help. This is a lot clearer to me.



I got one question, if cos(x)=1 and our solution interval is 0≤ x≤ 4pi would then cos(x) be the solution since x=0+2kpi

No! x=0+2kpi for k = 0 +/- 1, +/- 2, +/- 3 ,... So x = 0, +/- 2pi, +/- 4pi, +/- 6pi, .... The problem is that MANY of those results are not in [0, 4pi]. The correct answer is {0, 2pi, 4pi}

 

[Kalen12]

I too suspect that (0,pi) is not correct. I bet the problem stated that it included at least one of the end points.
In solving sin x = -1/2 on (0,pi) you really did not do the work which you did. As pointed out, if sin x <0, the x is in quad III or IV and NO angles in those quadrants are in (0,pi).

No, it's not a mistake it specifically says (0,pi).

No! x=0+2kpi for k = 0 +/- 1, +/- 2, +/- 3 ,... So x = 0, +/- 2pi, +/- 4pi, +/- 6pi, .... The problem is that MANY of those results are not in [0, 4pi]. The correct answer is {0, 2pi, 4pi}

Thanks, that's a bit clearer.