OK, I have a couple of math questions which I can't do. If you don't guess it, this is Year 12 math... I hope you can help. I really don't have much time to go and ask the teacher (I already have about 300 questions :P)

In the following two questions, you have to solve for x across [0, 2pi)

2sin2x = 1

I only obtained two answers ( pi/12 and its equivalent (13*pi)/12)

However, the book obtained FOUR. It had my answers, as well as (5*pi)/12 and (17*pi)/12

How could it get these two?! They are in the wrong quadrants...

The next question...

5cos(2x-(pi/6))=2

There are four answers again! And this question is too complex for me to even get two...

Can anyone help me, please?

How could it get these two?! They are in the wrong quadrants...

2 sin 2x = 1

sin 2x = ½

2x = arcsin ½

x = ½ arcsin ½

The sine of 2x must equal one-half, however, keeping x within the range [0, 2π] can yield four answers if 2x lies within the range [0, 4π].

i.e. 2x = π/6, 5π/6, 13π/6 and 17π/6.

x lies within the range [0, 2π] for all of the above.

2sin2x = 1

Therefore sin 2x = 1/2

2x = ..., pi/6, 5 pi/6, 13 pi/6, 17 pi/6, ...

and so, listing all answers between 0 and 2 pi, we have:

x = pi/12, 5 pi/12, 13 pi/12, 17 pi/12

5cos(2x-(pi/6))=2

cos (2x - pi/6) = 2/5

This has no easily-found exact solutions when you take the inverse cosine. You'll need to use a calculator.

Try solving it using graphics if you find it confusing to calculate those variables, it may help (at least in my case, which is sadly also my downfall... I failed my algebraic/calculus math subject twice in a row....)

For

0.4 = cos (2*(X - pi/12)),

where

0 < X =< 2pi,


X ≈

3291pi/12288 = 1097pi/4096,

11045pi/12288,

15579pi/12288 = 5193pi/4096, and

23333pi/12288.



Those are only approximate solutions, but I seriously doubt one can get much closer.

In the following two questions, you have to solve for x across [0, 2pi)

Note that this means you have to solve for 2x across [0, 4pi], so you're actually traversing each quadrant twice.

You might find it easier to substitute a for 2x, then solve for a across [0,4pi].

5cos(2x-(pi/6))=2

A substitution for the term in the trig function makes things easier

a = 2x - pi/6
0 <= x <= 2pi
-pi/6 <= a <= (4pi - pi/6)

5.cos(a) = 2
cos(a) = 2/5 = 0.4
a = acos(0.4)

There are four solutions for a in the range -pi/6 to 4pi - pi/6
For each solution, use x = (a + pi/6)/2 to find x

The first question is easy. Just add 2pi to the other two angles...

I still don't understand the second question.

I get the first two angles. But WHAT do I add to them to get the other two angles???

5cos(2x-(pi/6))=2 so cos(2x-(pi/6))= 2/5.

My calculator gives arccos(2/5)= 1.159 (approx) but if I take a quick glance at the graph of y= cos(x) I see that 2pi- 1.159= 5.124 also satisfies cos(x)= 2/5.
Indeed, 2pi+ 1.159= 7.442 and 4pi- 1.159= 11.407 also satisfy cos(x)= 2/5. Those are outside of [0,2pi) but THAT DOESN'T MEAN X IS!

Taking 2x- pi/6= 1.159 I get 2x= 1.159+ pi/6= 1.682 or x= 0.841
Taking 2x- pi/6= 5.124 I get 2x= 5.124+ pi/6= 5.647 or x= 2.824
Taking 2x- pi/6= 7.442 I get 2x= 7.442+ pi/6= 7.966 or x= 3.983
Taking 2x- pi/6= 11.407 I get 2x= 11.407+ pi/6= 11.931 or x= 5.965

All of which are in [0, 2pi)

YAY!
It all make sense now!
Thanks guys and girls ;)

Everyone disregard this post. The person who needs this knows who they are ;)

5cos(2x-(pi/6))=2

cos(2x-(pi/6)) =2/5

2x-(pi/6) = cos-1(2/5)

2x-(pi/6) = 2pi - cos-1(2/5), 2pi - cos-1(2/5) + 2pi, cos-1(2/5), cos-1(2/5) + 2pi

2x = 2pi - cos-1(2/5) + (pi/6), 2pi - cos-1(2/5) + 2pi + (pi/6), cos-1(2/5) + pi/6, cos-1(2/5) + 2pi + pi/6

x = all those answers divided by two!