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math help alevel trigonometry

I got the answers 65,155,245,335
however the mark scheme says 65 and 155 only,
the question is the following as it does not allow me to paste it in:

solve for 0(less than or equal to)x(less than)360
sin(x-20)=1/(square of 2)
Reply 1
Original post by .,...
I got the answers 65,155,245,335
however the mark scheme says 65 and 155 only,
the question is the following as it does not allow me to paste it in:

solve for 0(less than or equal to)x(less than)360
sin(x-20)=1/(square of 2)

just for clarity:
Solve 0<=x<=360
where sin(x-20)=1/sqrt(2)
right?
Reply 2
It helps to see your working, but it should be easy to sub the two extra values into the original equation and spot where youre reasoning is incorrect - it sholud be obv from the usual sin curve.
(edited 3 months ago)
Reply 3
Original post by eenie_49
just for clarity:
Solve 0<=x<=360
where sin(x-20)=1/sqrt(2)
right?

0<=x<360
where sin(x-20)=1/squareroot(2)
Reply 4
Original post by .,...
0<=x<360
where sin(x-20)=1/squareroot(2)

i gotta say i just solved it myself and idky the markscheme ony specifies two values when there are 4 solutions...

EDIT OH WAIT I GET IT NOWWWWWWW
(edited 3 months ago)
Reply 5
Original post by mqb2766
It helps to see your working, but it should be easy to sub the two extra values into the original equation and spot where youre reasoning is incorrect - it sholud be obv from the usual sin curve.

I did sine inverse(1/sqrt2) which gave me 45
so x-20=45 and using the cast method i got
x-20=45,135,225,315
then i added 20 to all the values on the rhs
then x=65,155,245,335
Reply 6
Original post by eenie_49
i gotta say i just solved it myself and idky the markscheme ony specifies two values when there are 4 solutions...

EDIT OH WAIT I GET IT NOWWWWWWW

Can you explain it to me please
Reply 7
okay you imagine a sin() graph right (pull up a picture of it) you can see that ony 65 and 155 are positive for sin()! because after 180, its negative! sketch below v
Reply 8
does that exaplin it??
Reply 9
Original post by eenie_49
does that exaplin it??

ohhh so if the angle in the sin is is positive then i cant use the negative bit of the graph?
Reply 10
Original post by eenie_49
does that exaplin it??

also how would i use the cast method to do it
Reply 11
Original post by .,...
ohhh so if the angle in the sin is is positive then i cant use the negative bit of the graph?

yh your right! if you see the dotted line that represents 1/sqrt(2) it wouldnt cross 335 or 245 between 0 and 360!
Reply 12
Original post by eenie_49
yh your right! if you see the dotted line that represents 1/sqrt(2) it wouldnt cross 335 or 245 between 0 and 360!

Thank you for helping me but i use the cast method so in that meethos how would i make sure i dont get the wrong anser
Reply 13
Original post by eenie_49
i gotta say i just solved it myself and idky the markscheme ony specifies two values when there are 4 solutions...

EDIT OH WAIT I GET IT NOWWWWWWW

Its generally a good habit to get into to sub answers back into the original equation to check.
Reply 14
Original post by mqb2766
Its generally a good habit to get into to sub answers back into the original equation to check.

that is smart thank you,but do u maybe know how i can use the cast method to solve this
Reply 15
Original post by .,...
that is smart thank you,but do u maybe know how i can use the cast method to solve this

okay whatever i try i cant upload the picture so im going to try expain it to you instead:

1) draw your cast diagram, with your un-transformed sin().which mean you put 45 degrees on your diagram, in your sin() and your all
2) you know your first answer is x-20=45 right!
3) your next answer, your moving clockwise from the 0 degrees to the 45 degrees in the sin(), draw your arrow, here so 180-45=135
4) so now x-20=45, 135
5) you wont do any more angles, beacuse this would mean drawing in the negative of the cast diagram (for sin(), the top half is positive and he bottom half is negative)
6) now you solve x=65, 155
7) hope that helped!! ill keep trying to get a picture for you!!
Reply 16
]https://app.gemoo.com/share/image-annotation/589570685990305792?codeId=v6gA6j1kL9a9l&origin=imageurlgenerator&card=589570683423391744
edit: awh sorry its really not working !! maybe try the link?? sorry tho!!
(edited 3 months ago)
Reply 17
Original post by .,...
that is smart thank you,but do u maybe know how i can use the cast method to solve this

cast or the symmetry of the sin curve simply says you have an extra solution about 90 (positive y axis if the solution is positive) or about 270 (negative y axis if its negative). So here
x-20 = 45, 135
If youre unsure, just sketch the sin curve and roughly spot where the line 1/sqrt(2) intersects it, so 90+/-45. If it was -1/sqrt(2) it would be 270+/-45 as should be clear from the sketch.
(edited 3 months ago)

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