Understanding Trig equationsTrig identity proof helpTrig equation help pleaseConfused by textbook solution to trig problema trig questionGeneral solutions for trigonometry equationsSimple Trig Equations - Why is it Wrong to Cancel Trig Terms?Extracting Like Variable Nested in Trig FunctionsIf $tan t = frac14$ and the terminal point for t is in Quadrant III, find $sec t + cot t$.What is wrong with this way of solving trig equations?If $sinx=t, quad xin(frac3pi2,2pi),$ what is $tanx?$
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Understanding Trig equations
Trig identity proof helpTrig equation help pleaseConfused by textbook solution to trig problema trig questionGeneral solutions for trigonometry equationsSimple Trig Equations - Why is it Wrong to Cancel Trig Terms?Extracting Like Variable Nested in Trig FunctionsIf $tan t = frac14$ and the terminal point for t is in Quadrant III, find $sec t + cot t$.What is wrong with this way of solving trig equations?If $sinx=t, quad xin(frac3pi2,2pi),$ what is $tanx?$
$begingroup$
I understand why $sin(x)=fracsqrt32$ has two answers, $fracpi3$ and $frac2pi3$ but I don't understand why $tan(x)=1$ only has one solution (according to my book and other places I've looked online) $fracpi4$. Why isn't $frac5pi4$ also a solution given that tan is positive in Q1 and Q3 and at $frac5pi4$ both $sin$ and $cos$ are $-fracsqrt22$
Note that the interval is not given.
trigonometry
asked Mar 17 at 21:23
dstarhdstarh
1426
$endgroup$
|
show 1 more comment
$begingroup$
I understand why $sin(x)=fracsqrt32$ has two answers, $fracpi3$ and $frac2pi3$ but I don't understand why $tan(x)=1$ only has one solution (according to my book and other places I've looked online) $fracpi4$. Why isn't $frac5pi4$ also a solution given that tan is positive in Q1 and Q3 and at $frac5pi4$ both $sin$ and $cos$ are $-fracsqrt22$
Note that the interval is not given.
trigonometry
asked Mar 17 at 21:23
dstarhdstarh
1426
$endgroup$
1
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
1
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33
|
show 1 more comment
$begingroup$
I understand why $sin(x)=fracsqrt32$ has two answers, $fracpi3$ and $frac2pi3$ but I don't understand why $tan(x)=1$ only has one solution (according to my book and other places I've looked online) $fracpi4$. Why isn't $frac5pi4$ also a solution given that tan is positive in Q1 and Q3 and at $frac5pi4$ both $sin$ and $cos$ are $-fracsqrt22$
Note that the interval is not given.
trigonometry
asked Mar 17 at 21:23
dstarhdstarh
1426
$endgroup$
I understand why $sin(x)=fracsqrt32$ has two answers, $fracpi3$ and $frac2pi3$ but I don't understand why $tan(x)=1$ only has one solution (according to my book and other places I've looked online) $fracpi4$. Why isn't $frac5pi4$ also a solution given that tan is positive in Q1 and Q3 and at $frac5pi4$ both $sin$ and $cos$ are $-fracsqrt22$
Note that the interval is not given.
trigonometry
trigonometry
asked Mar 17 at 21:23
dstarhdstarh
1426
asked Mar 17 at 21:23
dstarhdstarh
1426
edited Mar 17 at 21:36
dstarh
asked Mar 17 at 21:23
dstarhdstarh
1426
asked Mar 17 at 21:23
dstarhdstarh
1426
asked Mar 17 at 21:23
dstarhdstarh
1426
1426
1
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
1
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33
|
show 1 more comment
1
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
1
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33
1
1
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
1
1
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So,
$$sin(x)=fracsqrt32$$
has solutions,
$$fracpi3+2pi k$$
or
$$frac2pi3+2pi k$$
for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
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active
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votes
$begingroup$
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So,
$$sin(x)=fracsqrt32$$
has solutions,
$$fracpi3+2pi k$$
or
$$frac2pi3+2pi k$$
for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
|
show 2 more comments
$begingroup$
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So,
$$sin(x)=fracsqrt32$$
has solutions,
$$fracpi3+2pi k$$
or
$$frac2pi3+2pi k$$
for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
|
show 2 more comments
$begingroup$
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So,
$$sin(x)=fracsqrt32$$
has solutions,
$$fracpi3+2pi k$$
or
$$frac2pi3+2pi k$$
for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
$endgroup$
If the interval is not given, you would be wise to assume it's from minus infinity to positive infinity.
Consequently you should give the general solution.
So,
$$sin(x)=fracsqrt32$$
has solutions,
$$fracpi3+2pi k$$
or
$$frac2pi3+2pi k$$
for integer values of $k$
Shall I leave you to do similar for the other equations in your question ?
Top Tip : Drawing a graph of the appropriate trigonometric function is helpful.
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
answered Mar 17 at 21:33
Martin HansenMartin Hansen
707114
707114
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
|
show 2 more comments
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
Ok since there are 2 general solutions to sin(x) then you list both but since it's only one technical solution for tan(x) you would only list the one? That makes more sense. Thanks
$endgroup$
– dstarh
Mar 17 at 21:35
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
$begingroup$
If the interval IS given for ex from $0 < x < 2pi$ then would both solutions be valid for the tan equation?
$endgroup$
– dstarh
Mar 17 at 21:37
1
1
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
The period of the $tan(x)$ graph is just $pi$ radians so you can typically compact it's general solutions more than for general solutions from the $sin(x)$ and $cos(x)$ graphs with their period of $2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:38
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
$begingroup$
Ahh that makes sense. For whatever reason my professor skipped graphing of tan and just focused on sin/cos
$endgroup$
– dstarh
Mar 17 at 21:39
1
1
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
$begingroup$
Just to be clear : $tan(x)=1$ has the two solutions $fracpi4$ and $frac5pi4$ in the interval $0<x<2pi$
$endgroup$
– Martin Hansen
Mar 17 at 21:41
|
show 2 more comments
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1
$begingroup$
$sin(x) = sqrt3/2$ has infinitely many solutions, as does $tan(x) = 0$. That being said, I don't think that $pi/3$ solves $tan(x) = 0$. Nor do I agree with your assertion that $sin(4pi/3) = cos(4pi/3) = -sqrt2/2$.
$endgroup$
– Xander Henderson
Mar 17 at 21:27
$begingroup$
Sorry meant $frac5pi4$ editing
$endgroup$
– dstarh
Mar 17 at 21:29
$begingroup$
Do you mean $tan(x) = 1$? Regardless, both equations have infinitely many solutions as @XanderHenderson mentioned.
$endgroup$
– Klaus
Mar 17 at 21:30
$begingroup$
You ask "why isn't $frac5pi4$ also a solution," yet $x=frac5pi4$ is indeed a solution: $tan(frac5pi4)=1$
$endgroup$
– Noble Mushtak
Mar 17 at 21:33
1
$begingroup$
Yes, edited to fix. So in this case since there is no interval given it's assumed that it's just $fracpi4 + pi*n$ ?
$endgroup$
– dstarh
Mar 17 at 21:33