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.