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Trig

the limits are

0∘≤x<180∘0^\circ \leq x < 180^\circ

cos(3x-10)=-0.4

i did this for the limits, is this right?


0∘≤3x−10<180∘0^\circ \leq 3x-10 < 180^\circ


−10∘≤3x<170∘-10^\circ \leq 3x < 170^\circ


−30∘≤x<510∘-30^\circ \leq x < 510^\circ
(edited 7 years ago)
Reply 1
Avatar for Zacken
Zacken
22
Original post by thefatone
i did this for the limits, is this right?


Nopes. If 0<3x−10<1800 < 3x-10<180 then +10<3x<180+10+10 < 3x < 180 + 10
Reply 2
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
Nopes. If 0<3x−10<1800 < 3x-10<180 then +10<3x<180+10+10 < 3x < 180 + 10


but surely if i have an answer of -10 (just an example) then if i add 10 then divide by 3 then the answer would be in the limit of

0∘≤x<180∘0^\circ \leq x < 180^\circ


wouldn't it?
Reply 3
Avatar for Zacken
Zacken
22
Original post by thefatone
but surely if i have an answer of -10 (just an example) then if i add 10 then divide by 3 then the answer would be in the limit of

0∘≤x<180∘0^\circ \leq x < 180^\circ


wouldn't it?



I dunno what you're saying, but I advise that you look up solving linear equations.

It's like you're telling me that 3x−10=0  ⟺  3x=−103x-10 = 0 \iff 3x = -10
Reply 4
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thefatone
OP
6
Original post by Zacken
I dunno what you're saying, but I advise that you look up solving linear equations.

It's like you're telling me that 3x−10=0  ⟺  3x=−103x-10 = 0 \iff 3x = -10


not quite

i'm trying to say that if you have 3x-10=0 then on the left side then on the right side you must add 10

so this
0∘≤3x−10<180∘0^\circ \leq 3x-10 < 180^\circ

the answer can be 10 less because you add 10 when you get the final answers
thus
−10∘≤3x<170∘-10^\circ \leq 3x < 170^\circ
Reply 5
Avatar for Zacken
Zacken
22
Original post by thefatone
not quite

i'm trying to say that if you have 3x-10=0 then on the left side then on the right side you must add 10

so this
0∘≤3x−10<180∘0^\circ \leq 3x-10 < 180^\circ

the answer can be 10 less because you add 10 when you get the final answers
thus
−10∘≤3x<170∘-10^\circ \leq 3x < 170^\circ


What?

When you add 10 you get:

Unparseable latex formula:

\displaystyle [br]\begin{align*}0 < 3x - 10 < 180 &\Rightarrow 0 + 10 < 3x -10 + 10 < 180 + 10\\& \Rightarrow 10 < 3x + 0 < 190 \\&\Rightarrow 10 < 3x < 190 \end{align*}

Reply 6
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
What?

When you add 10 you get:

Unparseable latex formula:

\displaystyle [br]\begin{align*}0 < 3x - 10 < 180 &\Rightarrow 0 + 10 < 3x -10 + 10 < 180 + 10\\& \Rightarrow 10 < 3x + 0 < 190 \\&\Rightarrow 10 < 3x < 190 \end{align*}



i see, so then i multiply by 3 after that?
Reply 7
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Zacken
22
Original post by thefatone
i see, so then i multiply by 3 after that?


DIVIDE by 33.
Reply 8
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thefatone
OP
6
Original post by Zacken
DIVIDE by 33.


this is very confusing... but then when i get my values for
cos(3x-10)=-0.4

i wouldn't include like values any bigger than 63.3°
Reply 9
Avatar for Zacken
Zacken
22
Original post by thefatone
this is very confusing... but then when i get my values for
cos(3x-10)=-0.4

i wouldn't include like values any bigger than 63.3°


Yes, precisely. (also you wouldn't include values smaller than 3.33)
Reply 10
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
Yes, precisely. (also you wouldn't include values smaller than 3.33)


wat... that's not working with my logic and "correct" answers

here have the question and mark scheme

https://a44694f152ee562e38a9454969d2b9a549a03543.googledrive.com/host/0B1ZiqBksUHNYOTR1Zk42aXdIazg/05%20Silver%201%20-%20C2%20Edexcel.pdf
question 4

mark scheme is in the pages far down
Reply 11
Avatar for Zacken
Zacken
22
Original post by thefatone
wat... that's not working with my logic and "correct" answers

here have the question and mark scheme

https://a44694f152ee562e38a9454969d2b9a549a03543.googledrive.com/host/0B1ZiqBksUHNYOTR1Zk42aXdIazg/05%20Silver%201%20-%20C2%20Edexcel.pdf
question 4

mark scheme is in the pages far down


Urgh, I read your post wrong. I assumed your inequality was right.

We have: 0<x<180⇒3×0<3x<3×180⇒0<3x<540⇒−10<3x−10<5300 < x < 180 \Rightarrow 3 \times 0 < 3x < 3 \times 180 \Rightarrow 0 < 3x < 540 \Rightarrow -10 < 3x-10 < 530
Reply 12
Avatar for thefatone
thefatone
OP
6
Original post by Zacken
Urgh, I read your post wrong. I assumed your inequality was right.

We have: 0<x<180⇒3×0<3x<3×180⇒0<3x<540⇒−10<3x−10<5300 < x < 180 \Rightarrow 3 \times 0 < 3x < 3 \times 180 \Rightarrow 0 < 3x < 540 \Rightarrow -10 < 3x-10 < 530


so you multiply first then takeaway the 10???
Reply 13
Avatar for Zacken
Zacken
22
Original post by thefatone
so you multiply first then takeaway the 10???


Yeah.
Reply 14
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thefatone
OP
6
Original post by Zacken
Yeah.


Thanks for clearing that up
Reply 15
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Zacken
22
Original post by thefatone
Thanks for clearing that up


No problem.

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