General Solutions to Trigonometric Equations (2 Viewers)

[bored of sc]

This is from Maths In Focus, Exercise 6.20.

Here's my working:
5) sin2x = cosx
2sinxcosx = cosx
2sinxcosx - cosx = 0
cosx (2sinx - 1) = 0
cos x = 0
x = 90^o
2sinx - 1 = 0
sinx = 1/2
x = 30^o
Therefore,
x = 180^on + (-1)ⁿ30^o for sinx = 1/2 <--- That's correct.
x = 360^on + 90^o for cosx = 0 <-- That's incorrect.
My teacher told me: the fact that you have both the positive and negative cases and/or you have 90⁰ that 180^on + 90⁰ is sufficent since it covers all four quadrants, thus all possible answers.
But I need more explanation for this particular question.


Also...
7) sin @ = 0
@ = 0^o, 180^o, -180^o
@ = 180^on + (-1)ⁿ 0^o
@ = 180^on <-- That's half correct.


Answers???
5) x = 180^o x n + (-1)ⁿ x 30^o, 180^o x n + 90^o
7) @ = + 180^o x n

 

[independantz]

for 5 i think it is a mistake as the genral solution of cos is always 2(pie)n+/- (alpha)

and for 7 your answer is equivalent to their answer as n is an integer, not a positive integer.

Although it's been a while since i done this stuff so correct me if i'm wrong.

 

[bored of sc]

For question 7, that makes a lot of sense. Thank you for your clear powers of logic.


But for question 5 my teacher alerted me to the fact of certain conditions making the 360 part of the general formula 180. In the case of question 5, since x is 90, -270, -90, 270 this means 90 and -270 are in the same position (on the boarder of the 1st and second quadrants) and -90 and 270 are in the same position (on the boarder of the 3rd and fourth quadrants). Thus, a 90 degree turn in either a clockwise and anti-clockwise direction (180^o all up) covers all the general solutions.

Anyone care to verify? Criticise? Enlighten? Clarify? HELP!

 

[tommykins]

回复: Re: General Solutions to Trigonometric Equations

For question 7, that makes a lot of sense. Thank you for your clear powers of logic.


But for question 5 my teacher alerted me to the fact of certain conditions making the 360 part of the general formula 180. In the case of question 5, since x is 90, -270, -90, 270 this means 90 and -270 are in the same position (on the boarder of the 1st and second quadrants) and -90 and 270 are in the same position (on the boarder of the 3rd and fourth quadrants). Thus, a 90 degree turn in either a clockwise and anti-clockwise direction (180^o all up) covers all the general solutions.

Anyone care to verify? Criticise? Enlighten? Clarify? HELP!

90 = pi/2
270 = 3pi/2
-90 = -pi/2
-270 = -3pi/2

From this you can see that the general solution is pi/2 +- kpi where k = ..-2,-1,0,1,2,3,....k

 

[bored of sc]

Re: 回复: Re: General Solutions to Trigonometric Equations

90 = pi/2
270 = 3pi/2
-90 = -pi/2
-270 = -3pi/2

From this you can see that the general solution is pi/2 +- kpi where k = ..-2,-1,0,1,2,3,....k

Edit: pi = 180 right?

So x = 180k +- 90 where k is any integer.

 

[tommykins]

回复: Re: 回复: Re: General Solutions to Trigonometric Equations

Pi? Ahhhh! What's pi? It's not the irrational number equal to any circle's circumference/diameter is it?

it is. you work in radians in year 12, get used to it.

pi = 180* (pronounced pie however)

 

[bored of sc]

Re: 回复: Re: 回复: Re: General Solutions to Trigonometric Equations

it is. you work in radians in year 12, get used to it.

pi = 180* (pronounced pie however)

Okay. Thanks a tonne for all your help.

 

[tommykins]

Re: 回复: Re: General Solutions to Trigonometric Equations

Edit: pi = 180 right?

So x = 180k +- 90 where k is any integer.

x = pi/2 +- kpi = 90 +- k180 where k = ...-2,-1,0,1,2,3...

You need to specify that it's infinitely negative as well.

 

[Trebla]

This is from Maths In Focus, Exercise 6.20.

Here's my working:
5) sin2x = cosx
2sinxcosx = cosx
2sinxcosx - cosx = 0
cosx (2sinx - 1) = 0
cos x = 0
x = 90^o
2sinx - 1 = 0
sinx = 1/2
x = 30^o
Therefore,
x = 180^on + (-1)ⁿ30^o for sinx = 1/2 <--- That's correct.
x = 360^on + 90^o for cosx = 0 <-- That's incorrect.
My teacher told me: the fact that you have both the positive and negative cases and/or you have 90⁰ that 180^on + 90⁰ is sufficent since it covers all four quadrants, thus all possible answers.
But I need more explanation for this particular question.

When you solve: cos x = 0, the solutions are: x = ±90º, ±270º, ±450º, ±630º, ±810º, ±990º etc...

We can write them as:
x = ±90º, ±(180 + 90)º, ±(360 + 90)º, ±(540 + 90)º, ±(720 + 90)º, ±(900 + 90)º etc...
Implying:
x = ±90º, 180º(±1) ± 90º, 180º(±2) ± 90º, 180º(±3) ± 90º, 180º(±4) ± 90º, 180(±5) ± 90º etc...
The first terms of are all multiples of 180º, thus the general solution is: 180ºn ± 90º for integer n

We can also write them as:
x = ±90º, ±(360 - 90)º, ±(360 + 90)º, ±(720 - 90)º, ±(720 + 90)º, ±(1080 - 90)º etc...
Implying:
x = ±90º, 360º(±1) ± 90º, 360º(±2) ± 90º, 360º(±3) ± 90º etc...
The first terms of are all multiples of 360º, thus the general solution is: 360ºk ± 90º for integer k.

Both are correct. If you add 90º to 180º, it is equivalent to subtracting 90º from 360º.

Also, 360º is a MULTIPLE OF 180º so the two versions are equivalent except with one integer being twice of the other (i.e. n = 2k, which are still arbitrary integers)

 

[bored of sc]

When you solve: cos x = 0, the solutions are: x = ±90º, ±270º, ±450º, ±630º, ±810º, ±990º etc...

We can write them as:
x = ±90º, ±(180 + 90)º, ±(360 + 90)º, ±(540 + 90)º, ±(720 + 90)º, ±(900 + 90)º etc...
Implying:
x = ±90º, 180º(±1) ± 90º, 180º(±2) ± 90º, 180º(±3) ± 90º, 180º(±4) ± 90º, 180(±5) ± 90º etc...
The first terms of are all multiples of 180º, thus the general solution is: 180ºn ± 90º for integer n

We can also write them as:
x = ±90º, ±(360 - 90)º, ±(360 + 90)º, ±(720 - 90)º, ±(720 + 90)º, ±(1080 - 90)º etc...
Implying:
x = ±90º, 360º(±1) ± 90º, 360º(±2) ± 90º, 360º(±3) ± 90º etc...
The first terms of are all multiples of 360º, thus the general solution is: 360ºk ± 90º for integer k.

Both are correct. If you add 90º to 180º, it is equivalent to subtracting 90º from 360º.

Also, 360º is a MULTIPLE OF 180º so the two versions are equivalent except with one integer being twice of the other (i.e. n = 2k, which are still arbitrary integers)

Wow! That makes perfect sense. Cheers for wasting your time on me.