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lil_me
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#1
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#1
Solve algebraically the equation

*3sinx-1=0 0<x<360

* stands for squareroot sign
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madhapper
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rt3 sinx = 1

3sinx = 1

sinx = 1/3

x = sin^-1 (1/3)

so basically find all x between 0 and 360 such that sinx = 1/3
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Gaz031
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For the last part you'll want to utilise the ASTC quadrant rule.
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yazan_l
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rt3 sinx = 1

3sinx = 1
i cant get what u mean here
did u multiply by sqrt[3] or squared the equation?because either ways, it's not right!


Edit: unless the equation is meant to mean :
rt[3sinx] - 1 =0
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yazan_l
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*3sinx-1=0
what does this mean?
is it: rt[3] sinx - 1 =0
or rt[3sinx] - 1 = 0
or rt[3sinx-1] = 0 ( i dont think it's this one!)
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yazan_l
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if it was :
rt[3] sinx - 1 =0
then:
rt[3] sinx = 1
sinx = 1/rt3
x = arcsin (1/rt3)

if it was :
rt[3sinx] - 1 = 0
then:
rt[3sinx] = 1
3sinx = 1
sinx = 1/3
x = arcsin (1/3) [ just as madhapper did]!
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lil_me
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(Original post by yazan_l)
i cant get what u mean here
did u multiply by sqrt[3] or squared the equation?because either ways, it's not right!


Edit: unless the equation is meant to mean :
rt[3sinx] - 1 =0
yea i think it is rt[3sinx] - 1 =0 so wot do u do with the rt sign??

Also yea u have to do the ASCT thing to get the answers!
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yazan_l
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yea i think it is rt[3sinx] - 1 =0 so wot do u do with the rt sign??
i think that you know that rt[a] = a^(1/2)
so rt[3sinx] = 1
is : (3sinx)^(1/2) = 1
square both sides:
[(3sinx)^(1/2)]² = 1²

and u know that (a^n)^m = a^(mn)
so: [(3sinx)^(1/2)]² = 1²
is : [(3sinx)^(2/2)] = 1
so it is (3sinx)^1 = 1²
which is 3sinx = 1
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lil_me
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#9
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(Original post by yazan_l)
i think that you know that rt[a] = a^(1/2)
so rt[3sinx] = 1
is : (3sinx)^(1/2) = 1
square both sides:
[(3sinx)^(1/2)]² = 1²

and u know that (a^n)^m = a^(mn)
so: [(3sinx)^(1/2)]² = 1²
is : [(3sinx)^(2/2)] = 1
so it is (3sinx)^1 = 1²
which is 3sinx = 1
Kk thanx
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