# 0 <= x <= 2pi

1. Aug 21, 2008

### fr33pl4gu3

2 cos( 2 x ) +sin( x ) = 2
sin(x)= 2 -2 cos (2x)
= 2 (1 - cos (2x))
= 2 (2 sin 2x)
= 4 sin2x
0= 4sin2x - sinx
= (sin x)(4sin2x-1)

x = 0 and 0.25 (correct to 2 decimal place)

This is only partially correct, what's wrong??

2. Aug 21, 2008

### NoMoreExams

I assume in your pre-last step you mean $$4sin^{2}(x) - sin(x) = 0 \Rightarrow sin(x) \left( 4sin(x) - 1 \right) = 0$$. So you have to evaluate 2 possibilities, $$sin(x) = 0$$ and $$sin(x) = \frac{1}{4}$$. The first one gives you x = 0 AND $$\pi$$ (and actually $$2\pi$$ but I will assume you meant $$0 \leq x < 2 \pi$$)now think about where $$sin(x) = \frac{1}{4}$$

3. Aug 21, 2008

### HallsofIvy

Staff Emeritus
For small values of x (in radians), sin(x) is approximately equal to x so, yes, to two decimal places, arcsin(.25)= .25. But it is also true that $sin(\pi- x)= sin(x)$ so just as sin(x)= 0 gives both x= 0 and $x= \pi$, so sin(x)= .25 has two solutions between 0 and $2\pi$.

4. Aug 21, 2008

### fr33pl4gu3

sin x = 1/4
x= 0.25 (correct to 2 decimal place)

5. Aug 21, 2008

### fr33pl4gu3

But the quiz system note that got 5 distinct solution, how is it then?

6. Aug 21, 2008

### fr33pl4gu3

2 cos( 2 x ) +sin( x ) = 2

how to turn sin(x) into cosx??

7. Aug 21, 2008

### NoMoreExams

I just showed you how $$sin(x) = 0 \Rightarrow x = 0, \pi, 2 \pi$$. You were also told that $$sin(x) = \frac{1}{4}$$ will have 2 solutions. 3 + 2 solutions = 5 solutions.

8. Aug 21, 2008

### NoMoreExams

Why would you? Rewrite cos(2x) as some function of sin(x).

9. Aug 21, 2008

### HallsofIvy

Staff Emeritus
Of course! I am so used to $0\le x< 2\pi$ I didn't even notice that $2\pi$ was included!

10. Aug 21, 2008

### fr33pl4gu3

so, the answer would be: 0, 0.25, 1, 0.85, 2

Last edited: Aug 21, 2008
11. Aug 21, 2008

### NoMoreExams

You were just told that $$\pi$$ and $$2 \pi$$ were answers as well... do you understand why for example $$sin(x) = \frac{1}{4}$$ would have 2 solutions on $$0 \leq x \leq 2 \pi$$

12. Aug 21, 2008

### fr33pl4gu3

Not really, though. This is my guess, it has 2 solution because one is + and the other is -. or one is a quater and the other is 3 and a quater.

13. Aug 21, 2008

### NoMoreExams

...I'm not sure what you said but no. Think of a simpler situation $$sin(x) = \frac{\sqrt{2}}{2}$$, how many solutions does that have for $$0 \leq x \leq 2 \pi$$

14. Aug 21, 2008

### fr33pl4gu3

0 <= 1/4pi <= pi <= 5/4pi <= 2pi

15. Aug 21, 2008

### NoMoreExams

$$\frac{5 \pi}{4}$$ is in the 3rd quadrant where sin(x) is negative, so that is def. not one of the solutions.

16. Aug 21, 2008

### fr33pl4gu3

(5/4)pi is in the 3rd quadrant where sin(x) is negative, so that is def. not one of the solutions.

so, the other solution should be 3/4 pi which is on the second quadrant, and sin is +value.

17. Aug 21, 2008

### NoMoreExams

Yes, you can check that that is indeed the solution by plugging it back into your equation. Similarly $$sin(x) = \frac{1}{4}$$ will have 2 solutions. You found one already which you said was $$x = \frac{1}{4}$$ (it's actually 0.252680255 I believe but close enough), now find the 2nd one since I hopefully just convinced you that there should be 2. Note that you won't be able to do this by hand, but you do know that it will exist in the 2nd quadrant.

18. Aug 21, 2008

### fr33pl4gu3

so, how to do it by graph??

19. Aug 21, 2008

### fr33pl4gu3

so, how do i write pi and 2pi in value??

pi =1; 2pi = 2??

20. Aug 21, 2008

### fr33pl4gu3

the value on the second quandrant, is it x = -0.25?? but there is no negative value right. if one of the solution is 0.25 on the first quadrant, then the solution on the second quadrant should be -0.25, because the width it is the same. Correct??