Math Destroyer problem :(

[shyfox9p]

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Q: which of the following is a solution to the trigonometric equation 2sin(x)+1=csc(x)
Solution: 30 degrees

In the solution, it rewrites the equation as (2sin(x)-1)(sin(x)+1)=0
Then it says, "setting the first factor equal to zero, we get 2sin(x)-1=0, or sin x=1/2

If you do the same thing, but to the second factor, you get a different answer. I'd really appreciate it if someone could explain why we must use the first factor and not the second one.

Thanks

 

[djwon1]

I am not 100% positive but if you use the second factor, the answer comes out as 270 degrees.
However, if you plug 270 into csc x, which is the same as 1/sinx, you get zero in the denominator.

Or you might have range for x or 270 might not even be one the choices.

 

[shyfox9p]

Thanks.

 

[7amsho]

do they give you a specific range for the answer? because I got 270 and 30 and here is how I did it:
2*sin(x)+1=1/sin(x)
2*sin^2(x)+sin(x)=1 (I just multiplied by sin(x))
2*sin^2(x)+sin(x)-1=0 (now factor)
(2*sin(x)-1)*(sin(x)+1)=0 (this is like a*b=0 so either a=0 or b=0)
2*sin(x)-1=0 or sin(x)+1=0
sin(x)=1/2 or -1

if you know your trig table well, sin(x)=1/2 when it is 30 deg and sin(x)=-1 when it is 270 deg..

I hope this helps

 

[virajpatel]

^^^^ great explanation

 

[shyfox9p]

They only gave one range. I wish Dr. Romano would have given both answers. I really appreciate the explanation 7amsho!