trig equation qn (1 Viewer)

Masaken

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need help on what I may have done wrong with this qn? the answers say the solutions are only plus minus pi/2 + k2pi (which makes no sense to me because sin(-90) = -1? or are the answers just scuffed? thanks in advance

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5uckerberg

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need help on what I may have done wrong with this qn? the answers say the solutions are only plus minus pi/2 + k2pi (which makes no sense to me because sin(-90) = -1? or are the answers just scuffed? thanks in advance

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With the solution I believe that they took the negative root as well which can be justified through similar working.
 

Masaken

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With the solution I believe that they took the negative root as well which can be justified through similar working.
oh okay, what kind of working would you need to put in? also, the solutions only list plus minus pi/2 + k2pi, are the solutions for sinx = -1/2 correct as well? i have no idea why they wouldn't be correct (or why they would be solutions for -1/2 would be omitted in the first place, unless i've read or done the qn wrong?)
 

5uckerberg

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I believe that instead of the question is and next step you just have




. This I believe is the best I can come up with, I mean we can experiment with the values here.
 

Masaken

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i mean it's strange that the sign in the question would have to be changed because i have no idea how you would incorporate that into the working out but i guess the question is probably scuffed itself :/ thanks for the help
 

Masaken

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I believe that instead of the question is and next step you just have




. This I believe is the best I can come up with, I mean we can experiment with the values here.
i re-did the question based off of this and ended up getting the appropriate solutions, it didn't seem obvious to me before, but rather than simplifying 1 + cot^2(x) to csc^2(x), i changed cot to cos/sin, then with some simplifying you end up with cos2x + |sinx| = 0, then you solve considering two separate cases where sinx > 0 (or equal to as well) or sinx < 0, then ended up getting the correct answers (just plus minus pi/2 + k2pi), thanks for the help :)
 

5uckerberg

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i re-did the question based off of this and ended up getting the appropriate solutions, it didn't seem obvious to me before, but rather than simplifying 1 + cot^2(x) to csc^2(x), i changed cot to cos/sin, then with some simplifying you end up with cos2x + |sinx| = 0, then you solve considering two separate cases where sinx > 0 (or equal to as well) or sinx < 0, then ended up getting the correct answers (just plus minus pi/2 + k2pi), thanks for the help :)
You can actually rule out the solution because the equation does not match.
 

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I got -pi/6+2npi, where n is an integer and pi/2+2npi, where n is an integer
 

5uckerberg

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Was it incorrect? I got same answers as you did
i re-did the question based off of this and ended up getting the appropriate solutions, it didn't seem obvious to me before, but rather than simplifying 1 + cot^2(x) to csc^2(x), i changed cot to cos/sin, then with some simplifying you end up with cos2x + |sinx| = 0, then you solve considering two separate cases where sinx > 0 (or equal to as well) or sinx < 0, then ended up getting the correct answers (just plus minus pi/2 + k2pi), thanks for the help.

This is the key part.
 

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Oh alr so we took +- from that so therefore lcscxl?
I'll see what I get w that
 

Masaken

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Bro when tf did absolute values come in πŸ’€
sorry if the handwriting is completely illegible, lmk if you can't understand it
1672984227693.png
the key to getting the absolute value is what happens when you root sin^2(x), because you get both -sinx and sinx as solutions
 

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Yeah I got +- pi/2 +2npi, but theres also the other two solutions for pi/6+2npi tho...
 

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do we disregard it as we would also need to pick up the (pi-(+-1/2)) which would lead to an additional 2 solutions?
 

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