# Stuck on Trigonometry equations problemWatch

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#1
Hello.

I can't seem to work out how to do this. Any guidance/help is appreciated.

Find the solutions of the following equation for the domain 0 ≦ θ ≦ 360°

2 = + 1

Where do i start? Really stuck
0
5 years ago
#2
(Original post by Sir Phillip Jones)

Where do i start?
Hint: You have a quadratic in disguise.
0
5 years ago
#3
(Original post by Sir Phillip Jones)
Hello.

I can't seem to work out how to do this. Any guidance/help is appreciated.

Find the solutions of the following equation for the domain 0 ≦ θ ≦ 360°

2 = + 1

Where do i start? Really stuck

I would replace with an x and see what your equation looks like.
0
#4
(Original post by ghostwalker)
Hint: You have a quadratic in disguise.

I would replace with an x and see what your equation looks like.

2 = x + 1

What do i do from here? I'm not seeing the quadratic for some reason, to be able to simplify it
0
5 years ago
#5
(Original post by ghostwalker)
Hint: You have a quadratic in disguise.

I would replace with an x and see what your equation looks like.

These responses interest me - I do not see any disguise - surely it is just a quadratic - why would there be any need to change to a quadratic in x

IME this just means that people forget to solve the original equation - why not just solve the quadratic we are given
0
5 years ago
#6
(Original post by Sir Phillip Jones)
2 = x + 1

What do i do from here? I'm not seeing the quadratic for some reason, to be able to simplify it
Do you know that
0
#7
(Original post by TenOfThem)
Do you know that
Nope i didn't. Thank you. But i'm still confused about how to solve it and find the solutions for the domain. Not sure about the next steps.
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5 years ago
#8
(Original post by Sir Phillip Jones)
Nope i didn't. Thank you. But i'm still confused about how to solve it and find the solutions for the domain. Not sure about the next steps.

Which is the same as

Can you really not see the quadratic
1
#9
(Original post by TenOfThem)

Which is the same as

Can you really not see the quadratic

Ah yes i can see that now. Thanks very much. Replacing with x would then give:

- x - 1 = 0

Factorising that would give (2x + 1)(x - 1)

Changing it back to the original q and changing x again would give:

+1 = 0

&

- 1 = 0

Is this correct? How do i then proceed from here and simplify it to work out the solutions for the domain
0
5 years ago
#10
(Original post by TenOfThem)
These responses interest me - I do not see any disguise ...Can you really not see the quadratic
'nuf said!
0
5 years ago
#11
(Original post by Sir Phillip Jones)
Ah yes i can see that now.

+1 = 0

&

- 1 = 0

Is this correct? How do i then proceed from here and simplify it to work out the solutions for the domain

That is correct

I am sure that you can finish this

The first one gives

giving

etc
0
5 years ago
#12
(Original post by ghostwalker)
'nuf said!
but this was a bigger mis-understanding
0
#13
(Original post by TenOfThem)
That is correct

I am sure that you can finish this

The first one gives

giving

etc
Thanks again.

So my values for are -0.5 and 1

I then work out shift sin of both of these numbers on the calculator. I then get -30 and 90. Is this correct? And would these be the only two solutions satisfying the domain 0 ≦ θ ≦ 360°
0
5 years ago
#14
(Original post by Sir Phillip Jones)
Thanks again.

So my values for are -0.5 and 1

I then work out shift sin of both of these numbers on the calculator. I then get -30 and 90. Is this correct? And would these be the only two solutions satisfying the domain 0 ≦ θ ≦ 360°
Do not forget that those are solutions for you will need to find more solutions as, when you halve them, you will get solutions in the range

You should end up with 6 solutions

Oh and -30 would be no good anyway as that is outside of your region
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#15
(Original post by TenOfThem)
Do not forget that those are solutions for you will need to find more solutions as, when you halve them, you will get solutions in the range

You should end up with 6 solutions

Oh and -30 would be no good anyway as that is outside of your region
Can you explain to me how to find the other solutions. I get that it's for and for we just half -30 and 90. You then get -15 and 45. I also understand that -15 isn't counted as it's not in the range. but how do i see what the rest of the solutions are
0
5 years ago
#16
(Original post by Sir Phillip Jones)
Can you explain to me how to find the other solutions. I get that it's for and for we just half -30 and 90. You then get -15 and 45. I also understand that -15 isn't counted as it's not in the range. but how do i see what the rest of the solutions are

has the solutions for

210, 330, 210+360, 330+360, 210+720, 330+720, etc

I do not know which method your teacher has taught you to find these CAST, Graphs, Formulas

The first 4 will all halve to be in the range required
0
#17
(Original post by TenOfThem)

has the solutions for

210, 330, 210+360, 330+360, 210+720, 330+720, etc

I do not know which method your teacher has taught you to find these CAST, Graphs, Formulas

The first 4 will all halve to be in the range required
So it's 210, 330, 285, 345 and 2 others. I understand now how you worked these out. For the other two solutions, is one of them 45 and the other 135, 215 or 315?
0
5 years ago
#18
(Original post by Sir Phillip Jones)
So it's 210, 330, 285, 345 and 2 others. I understand now how you worked these out. For the other two solutions, is one of them 45 and the other 135, 215 or 315?
Check the others

Remember you ned an answer of 1 not -1
0
#19
(Original post by TenOfThem)
Check the others

Remember you ned an answer of 1 not -1
Sorry i'm confused again.

Are the remaining two answers 45 and (180+45) then?
0
5 years ago
#20
(Original post by Sir Phillip Jones)
Sorry i'm confused again.

Are the remaining two answers 45 and (180+45) then?
Do they give the correct answer - if so they are correct - you can check that
0
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