#i need help verifying the answer for this question
49 messages · Page 1 of 1 (latest)
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why do you think it’s incorrect?
im insanely fatigued and have been making stupid mistakes the entire day on this chapter so im not sure
i assume you have an answer key/solution book?
also congrats on helper of the week man!
to me, it looks like your method is fine.
i do not
i’ve has it for ~1 month, dw about it
the teacher forgot to unhide the answer in the slide it seems
i’ll double check it you made any errors though
ah ty
dark matter
ye and as its sin its always +ve
cuz 1st and 2nd quadrant
that's why i just put sin theta = x and made it a quadratic
to get 2 values
yeah, but don’t you think that could extend the solution set?
pi-x is not the same as x unless x is pi/2
clearly, that’s not the case
but isn't any trigo function (pi-x) always that function (x)
try that for cosine.
yeah, so not nevesssirl the same
my point is that you can expand the solution set
for the positive variant of x
no, it is
we are asked for values of theta, not x.
ooh right!
but suppose we get x = 1/2 and root3/2 so theta would be pi/6 and pi/3 so again we got 2 values
as the range of sin is defined we're only gonna get those values
not something like - pi/6 and such
this is what i meant by fatigued lol
i was just thinking of finding x rather than theta
lol
but i got lucky in this question
ah so this logic is correct rite?
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