0 <= x <= 2pi

[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)
The answer would be:

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

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

[NoMoreExams]

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)
The answer would be:

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

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

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}

[HallsofIvy]

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.

[fr33pl4gu3]

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

[fr33pl4gu3]

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

[fr33pl4gu3]

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

how to turn sin(x) into cosx??

[NoMoreExams]

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

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.

[NoMoreExams]

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

how to turn sin(x) into cosx??

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

[HallsofIvy]

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

[fr33pl4gu3]

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

[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

[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.

[NoMoreExams]

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.

...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

[fr33pl4gu3]

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

[NoMoreExams]

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

[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.

[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.

[fr33pl4gu3]

so, how to do it by graph??

[fr33pl4gu3]

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

pi =1; 2pi = 2??

[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??

[NoMoreExams]

Your questions don't make sense to me at all. How do you do it by graph? Graph f(x) = sin(x) and g(x) = 1/4, the 2 places they intersect on the interval [0, 2pi] is the solution set you are after. What do you mean pi = 1 and 2pi = 2? Pi is a constant (and so is 2pi obviously), how can they equal 1 and 2?. You find the value in the 2nd quadrant by estimation techniques which I am not sure you know yet.

[fr33pl4gu3]

the estimation technique got a rule, right, what is that rule, then??

[NoMoreExams]

You can expand sin(x) using MacLaurin and depending on how much accuracy you want, you can expand it to a finite number of terms, then figure out a "first" guess solution and then zero in on it.

[fr33pl4gu3]

i don't have many chances on guess, i only have 2 chance before my overall mark of that question became totally 0, if i got it correct, i get an overall mark of 0.2 out of 1.0, so, it's quite low already, instead, if got a definite technique that will get the answer, this'll be a great time.

[NoMoreExams]

If you need an exact answer, use a calculator or a CAS (Maple, Mathematica, etc. should do the trick).

[fr33pl4gu3]

ya, i got maple 12 but i don't know how to type in the equation to get the answer??

[NoMoreExams]

Maple has a great help section, you should learn to utilize it. For something as simple as this, I believe

solve(sin(x)=.25,x);

should do the trick but I am not sure if it would give the answer(s) in radians or degress. In any case you should know how to go between the 2 results.

[HallsofIvy]

Every point on the unit circle has coordinates (cos(t), sin(t)) where t is the angle the line from the origin to the point makes with the x- axis. In particular, if you draw a horizontal line through circle, you will get two points with the same 'y' value and so the same sin(t). From symmetry you should be able to see that if one is t, the other is \pi- t.