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Mathematics Challenge Marathon

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RuiAce:

--- Quote from: KoA on February 16, 2016, 05:09:45 pm ---Solution
--- End quote ---
The second part was done well.

However, whilst the first part was done correctly, it was not done in the most elegant manner:



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You may answer for 0°≤x≤360° if you have not been exposed to the radian measure of an angle.
SpoilerRequired knowledge: Preliminary Trigonometric Ratios

Happy Physics Land:
memes

jakesilove:

--- Quote from: Happy Physics Land on February 16, 2016, 07:08:10 pm ---memes

(Image removed from quote.)

--- End quote ---

Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

RuiAce:

--- Quote from: jakesilove on February 16, 2016, 07:10:26 pm ---Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

--- End quote ---

Yeah. The case that (sin(x)-cos(x))=0 has been disregarded in this case, and thus tan(x)=1 was ignored.

Happy Physics Land:

--- Quote from: jakesilove on February 16, 2016, 07:10:26 pm ---Gonna have to jump in there: Whilst it all looks like good maths, you divide by (sinx-cosx) between lines 2 and three. Whenever you divide by a term like that, you must explicitly state that (sinx-cosx) does not equal zero, and then using that equation figure out values for x that cannot exist (Maybe not in 2U, but since I know you do 4U...). This will likely change some of your final answers.

Jake

--- End quote ---

Ahhhhhh ok thank you so much Jake for pointing that out!!! I'm actually quite unaware of this detail and I guess its gonna be fatal in an exam (better jot this down).

ok so if l consider sinx - cosx cannot equal 0
then sinx cannot equal cos x
then tan x cannot equal 1
then x cannot equal pi/4 and that shouldnt really matter
but since we are dividing by cos x on both sides
cos x cannot equal 0
hence x cannot equal to pi/2 or 3pi/2

So the final answer for x would be x = 0, pi and 2pi?

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