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Further Pure 1 Triogonometry help please! watch

1. Find the general solutions, in radians, of this equation:

1 + sin(3x + 1) = 0.3

______

Having a lot of trouble with this one, can't get an answer that agrees with the mark scheme so if some lovely person could go through step by step for me that would be greatly appreciated?

Thank you,

Jack
2. What answer did you get? This is the kind of question where there are several correct answers so you may still be right.
3. Oh right, well I'm not really 100% sure how to work it out, because I get as far as

sin (3x + 1) = -0.7
3x + 1 = sin-1 -0.7
3x + 1 = -44.427

Then I'm not sure if I take -1 off both sides to make it:

3x = -45.427

Or whether I do it differently?

Jack
4. sin(3x+1)=-0.7

Let 3x+1 = u, so x = (u-1)/3 then

u = arcsin(-0.7), and

x = (arcsin(-0.7) - 1)/3.

You should obtain an answer that looks like , for integer values of k. (The -0.592 is to 3sf)

EDIT: Make sure you're working in radians, not degrees.
5. Thank you, yeah that's the answer in the mark scheme (one of them), but how do you get from...

x = (arcsin(-0.7) - 1)/3.

to

x = -0.592 + 2kpi
6. (Original post by Jack Dolan)
Thank you, yeah that's the answer in the mark scheme (one of them), but how do you get from...

x = (arcsin(-0.7) - 1)/3.

to

x = -0.592 + 2kpi
(Original post by marcusmerehay)
...
I also don't agree with that last step. It might help to sketch a graph of sin(u):

So you have sin(u)=-0.7. From there, you can get a general formula for u in terms of k, pi etc. Then once you have your general formula for u, use the fact that x=(u-1)/3 to get your general formula for x to be ((formula for u) - 1)/3
7. (Original post by Jack Dolan)
Thank you, yeah that's the answer in the mark scheme (one of them), but how do you get from...

x = (arcsin(-0.7) - 1)/3.

to

x = -0.592 + 2kpi
Compute arcsin(-0.7) = -0.77539749.... and substitute it in. Then recall that the sine graph repeats every 2pi radians.

EDIT: What ttoby said. I've made a fairly common error myself there by not looking at the graph.
8. Do I need to have done core 2 for this?

Jack
9. (Original post by Jack Dolan)
Do I need to have done core 2 for this?

Jack
No.

But you will have to solve cosx = -1/3, etc, for example.
10. It's all this radians stuff that is confusing me...

I'm teaching myself FP1 and got to chapter 5 and am baffled by some of these questions..

e.g. Compute arcsin(-0.7) = -0.77539749

?

Jack
11. (Original post by Jack Dolan)
It's all this radians stuff that is confusing me...

I'm teaching myself FP1 and got to chapter 5 and am baffled by some of these questions..

e.g. Compute arcsin(-0.7) = -0.77539749

?

Jack
arcsin is the same thing as .

By compute I meant do it on your calculator.

by definition. (+2kpi)
12. sin-1 -0.7 gives me -44.427.... Calc in the wrong mode I guess?

Jack
14. On the mark scheme it says 2npi / 3 - 0.5918c, and 2npi / 3 + 0.9723c.
Why is it /3, and where does the 0.9723c come from?

15. 1 + sin(3x + 1) = 0.3

sin(3x + 1) = -0.7

let 3x + 1 = X

so sin(X)=-0.7

X=-0.77539749

solve for other values by looking at its graph and calculating.

Draw the line y=-0.7 to see:

X= arcsin(-0.7)
16. Thank you for that, but I'm still not sure how you get to the answers:

x = 2npi / 3 - 0.5918c, and 2npi / 3 + 0.9723c.

Sorry :/
17. (Original post by Jack Dolan)
Thank you for that, but I'm still not sure how you get to the answers:

x = 2npi / 3 - 0.5918c, and 2npi / 3 + 0.9723c.

Sorry :/
So we have sin(u)=-0.7. We want to get a general formula for u, then from there get a general formula for x.

To get a general fomula for u, notice that (working in radians) the sin function has period 2pi and if you look at the graph I posted above, you'll see that if you take any two values 2pi apart, there are two solutions to sin(u)=-0.7 for u between those two values. So what we want to do is find those two solutions (u=a or b for example) then say that because sin has period 2pi, we know that a, a+2pi, a+4pi, b, b+2pi, b+4pi etc are all solutions to sin(u)=-0.7.

We can get one solution easily: just calculate sin^-1(-0.7) to get -0.775397497. This can be our 'a'. Now the next bit might be a bit harder if you haven't done C2 yet. Sketch a graph of y=sin(u) in radians and on the same axes sketch the line y=-0.7. Mark on that graph the point (-0.775397497, -0.7). This should be at an intersection between the curve and the straight line. Now notice that if you put your fnger on the point (pi, 0) and move your finger to the right by 0.775397497 then take the sin of the number you're on then you'll get -0.7. This is because of the symmetry of the curve. So you have two values fairly close to each other whose sin is -0.7. These are -0.775397497=a and pi+0.775397497 = 3.91699015 which we will call b. I have marked these on the graph below:

So now we can use the fact that sin has period 2pi. So we have solutions to sin(u)=-0.7 of u=a+2kpi where k can be any integer as well as solutions of u=b+2kpi where k is any integer (we can replace a and b with their actual values later on).

But the question is asking for x, not u. So we know x=(u-1)/3 so the two general formulas for x are and the the other general formula is .

Swapping round the order of the terms and rounding a bit gives x=2kpi/3 - 0.5918 and 2kpi/3 + 0.9723 which is equivilant to the answer in the book.

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Updated: December 11, 2010
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