Considering + and -

[__SB]

Hello, I was hoping if somebody could clear up a slightunderstanding error that I keep making whilst doing trigonometry. I will showtwo questions to illustrate my point.
Q1: Solve Sqr 17 cos(x + 14.0 ) = 2 * Sqr means squareroot (region is -180 < x < 180)
The method of doing this is fine; I obtain my two values of-75 , 46.9
However, I also considered that:
X + 14 = Cos^-1 (2 / -Sqr 17), because Sqr 17 can have a positiveor negative value
This gives me 2 more solutions which are incorrect. ( -133.1, 105)


Q2: Solve, for 0 <x < 180, the equation
tan( x + 60) * tan(x – 60) = 4sec^2 (x) – 3
Eventually I get to ;
x = tan^-1 (fourth root (4/12))
This time you have to considerthat the fourth root of 4/12 can be both positive and negative.
So basically I’m not sure whysometimes you consider a root can be both +/-, other times you just consider +.
I’d really appreciate if somebodycould help clarify this. Thanks in advance.

 

[stapel]

Sqr 17 can have a positiveor negative value

No. "The" square root is positive.

 

[Ishuda]

Hello, I was hoping if somebody could clear up a slightunderstanding error that I keep making whilst doing trigonometry. I will showtwo questions to illustrate my point.
Q1: Solve Sqr 17 cos(x + 14.0 ) = 2 * Sqr means squareroot (region is -180 < x < 180)
The method of doing this is fine; I obtain my two values of-75 , 46.9
However, I also considered that:
X + 14 = Cos^-1 (2 / -Sqr 17), because Sqr 17 can have a positiveor negative value
This gives me 2 more solutions which are incorrect. ( -133.1, 105)


Q2: Solve, for 0 <x < 180, the equation
tan( x + 60) * tan(x – 60) = 4sec^2 (x) – 3
Eventually I get to ;
x = tan^-1 (fourth root (4/12))
This time you have to considerthat the fourth root of 4/12 can be both positive and negative.
So basically I’m not sure whysometimes you consider a root can be both +/-, other times you just consider +.
I’d really appreciate if somebodycould help clarify this. Thanks in advance.

This is a question which arises quite often and the only answer available is "it is the convention". That is, given a = \(\displaystyle \sqrt{16}\), what is meant is a=4. That is \(\displaystyle \sqrt{a^2}\) = |a|. That is also why you see the solution to the quadratic equation
a x² + b x + c = 0, a \(\displaystyle \ne\) 0,
written as
x = \(\displaystyle \frac{-b\, \pm\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\).
Had you wanted only one solution then you would use only the appropriate, i.e the + or the - but not both, sign.
Thus in Q1 you are given the square root.

On the other had, when you are looking for a solution, what is the a such that a²=16, you need to get both solutions, a=\(\displaystyle \pm\)4.
In Q2 I suspect you introduced the square root in looking for the solution. Without going through the solution itself and assuming it is correct, I suspect what you may have arrived at was something like
tan²(x) = \(\displaystyle \frac{1}{\sqrt{3}}\)
which would be like that a²=16 above.

 

[__SB]

This is a question which arises quite often and the only answer available is "it is the convention". That is, given a = \(\displaystyle \sqrt{16}\), what is meant is a=4. That is \(\displaystyle \sqrt{a^2}\) = |a|. That is also why you see the solution to the quadratic equation
a x² + b x + c = 0, a \(\displaystyle \ne\) 0,
written as
x = \(\displaystyle \frac{-b\, \pm\, \sqrt{b^2\, -\, 4\, a\, c}}{2\, a}\).
Had you wanted only one solution then you would use only the appropriate, i.e the + or the - but not both, sign.
Thus in Q1 you are given the square root.

On the other had, when you are looking for a solution, what is the a such that a²=16, you need to get both solutions, a=\(\displaystyle \pm\)4.
In Q2 I suspect you introduced the square root in looking for the solution. Without going through the solution itself and assuming it is correct, I suspect what you may have arrived at was something like
tan²(x) = \(\displaystyle \frac{1}{\sqrt{3}}\)
which would be like that a²=16 above.

Thank you very much!! This clears it up perfectly.