need someone to check my work

[abel muroi]

i was given this problem 3 tan² x - 1 = 0 and was told to find all values of x (in degrees)

here is my work...


3 tan² x - 1 = 0
3 tan² x = 1
tan² x = 1/3
tan x = +- 1/(squareroot 3)

x = 30
x = -30


so 30 and -30 are my final answers

p.s i am not looking for a general solution, just solutions that are in degrees

 

[Steven G]

i was given this problem 3 tan² x - 1 = 0 and was told to find all values of x (in degrees)

here is my work...


3 tan² x - 1 = 0
3 tan² x = 1
tan² x = 1/3
tan x = +- 1/(squareroot 3)

x = 30
x = -30


so 30 and -30 are my final answers

p.s i am not looking for a general solution, just solutions that are in degrees

i am not looking for a general solution, just solutions that are in degrees. Are you suggesting that general solutions can't be in degrees? Why not? I suspect that you want the solutions within a specific range. If that is the case then include this information when you post your question. And yes 30 and -30 are solutions BUT maybe not in the range that the problem asked for.

 

[abel muroi]

i am not looking for a general solution, just solutions that are in degrees. Are you suggesting that general solutions can't be in degrees? Why not? I suspect that you want the solutions within a specific range. If that is the case then include this information when you post your question. And yes 30 and -30 are solutions BUT maybe not in the range that the problem asked for.

yes that was exactly what i meant. (i thought general solutions can only be in radians, but now i know it can be in degrees too).

then what range is the problem asking for?

 

[Steven G]

yes that was exactly what i meant. (i thought general solutions can only be in radians, but now i know it can be in degrees too).

then what range is the problem asking for?

Based on what you wrote (was told to find all values of x) there is no range. As I told you before I would find ALL solutions to the equation for one period and then add on multiples of the period length to your solutions. I am getting tired of saying this to you so I will do it for you.
One period of a tangent graph is between -90 and 90 (degrees). The length of this period (from -90 to 90) is 180. You found all solutions with this period. So ALL the solutions are 30 + 180n and -30 + 180n where n=0, +/-1, +/-2,...

 

[abel muroi]

Based on what you wrote (was told to find all values of x) there is no range. As I told you before I would find ALL solutions to the equation for one period and then add on multiples of the period length to your solutions. I am getting tired of saying this to you so I will do it for you.
One period of a tangent graph is between -90 and 90 (degrees). The length of this period (from -90 to 90) is 180. You found all solutions with this period. So ALL the solutions are 30 + 180n and -30 + 180n where n=0, +/-1, +/-2,...

ok i understand now, thank you.

one more question (i just want to be 100% sure about this) if i am asked to give the solutions between 0 - 360 degrees, does that mean that the only solution is 30 degrees? or is there another one that i am not seeing?

 

[Deleted member 4993]

ok i understand now, thank you.

one more question (i just want to be 100% sure about this) if i am asked to give the solutions between 0 - 360 degrees, does that mean that the only solution is 30 degrees? or is there another one that i am not seeing?

You have four solutions when the domain is restricted to [0,360°] and those are 30°, 150°, 210° & 330°

30 + 180 = 210 and

- 30 + 180 = 150

-30 + 360 = 330

 

[abel muroi]

You have four solutions when the domain is restricted to [0,360°] and those are 30°, 150°, 210° & 330°

30 + 180 = 210 and

- 30 + 180 = 150

-30 + 360 = 330

since the two solutions that i obtained from the problem are 30, and -30

does that mean that we subtract 30 from 180/360 BECAUSE we got a -30? or does it matter?

 

[ksdhart]

The answer to your question is yes, you subtract 30 from 180 because the 30 is negative. But it seems to me you're getting bogged down in the trigonometry parts of the problem and overthinking the process. If we go back to basic arithmetic for a moment, we notice that, by the commutative property of addition: -30+180n = 180n-30.

 

[Steven G]

since the two solutions that i obtained from the problem are 30, and -30

does that mean that we subtract 30 from 180/360 BECAUSE we got a -30? or does it matter?

-30 is the same angle as 330 (-30+360 =330)

 

[abel muroi]

The answer to your question is yes, you subtract 30 from 180 because the 30 is negative. But it seems to me you're getting bogged down in the trigonometry parts of the problem and overthinking the process. If we go back to basic arithmetic for a moment, we notice that, by the commutative property of addition: -30+180n = 180n-30.

yes sorry about that, i always have a tendency to over-think things a bit too much

 

[Otis]

In radians:

Pi*(n + 1/6) where n is an Integer

By the way, Jomo has confused the words 'domain' and 'range' in this thread.

 

[Steven G]

In radians:

Pi*(n + 1/6) where n is an Integer

By the way, Jomo has confused the words 'domain' and 'range' in this thread.

Interesting point you made! When I said range I was not referring to the word range in domain & range. I just meant interval. I will be more careful next time. Thanks for pointing this out to me.