# Hey help on this ?Watch

This discussion is closed.
#1
Solve algebraically the equation

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

* stands for squareroot sign
0
13 years ago
#2
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
0
13 years ago
#3
For the last part you'll want to utilise the ASTC quadrant rule.
0
13 years ago
#4
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
0
13 years ago
#5
*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!)
0
13 years ago
#6
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]!
0
#7
(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!
0
13 years ago
#8
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)]Â² = 1Â²

and u know that (a^n)^m = a^(mn)
so: [(3sinx)^(1/2)]Â² = 1Â²
is : [(3sinx)^(2/2)] = 1
so it is (3sinx)^1 = 1Â²
which is 3sinx = 1
0
#9
(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)]Â² = 1Â²

and u know that (a^n)^m = a^(mn)
so: [(3sinx)^(1/2)]Â² = 1Â²
is : [(3sinx)^(2/2)] = 1
so it is (3sinx)^1 = 1Â²
which is 3sinx = 1
Kk thanx
0
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