Mixing
Help with trigonometric equation $tan(x)+cot(2x)=1$
$begingroup$
I am trying to solve the equation $tan(x)+cot(2x)=1$.
It is clear that the equation has a solution $x=pi/4$ but I can't show that this is the only solution. Any ideas?
P.S. I am trying to solve the equation without using formulas for $2a$ angle
trigonometry
edited Jan 24 at 23:08
projectilemotion
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
$endgroup$
add a comment |
$begingroup$
I am trying to solve the equation $tan(x)+cot(2x)=1$.
It is clear that the equation has a solution $x=pi/4$ but I can't show that this is the only solution. Any ideas?
P.S. I am trying to solve the equation without using formulas for $2a$ angle
trigonometry
edited Jan 24 at 23:08
projectilemotion
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
$endgroup$
3
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
1
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48
add a comment |
$begingroup$
I am trying to solve the equation $tan(x)+cot(2x)=1$.
It is clear that the equation has a solution $x=pi/4$ but I can't show that this is the only solution. Any ideas?
P.S. I am trying to solve the equation without using formulas for $2a$ angle
trigonometry
edited Jan 24 at 23:08
projectilemotion
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
$endgroup$
I am trying to solve the equation $tan(x)+cot(2x)=1$.
It is clear that the equation has a solution $x=pi/4$ but I can't show that this is the only solution. Any ideas?
P.S. I am trying to solve the equation without using formulas for $2a$ angle
trigonometry
trigonometry
edited Jan 24 at 23:08
projectilemotion
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
edited Jan 24 at 23:08
projectilemotion
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
edited Jan 24 at 23:08
projectilemotion
11.5k62241
edited Jan 24 at 23:08
projectilemotion
11.5k62241
edited Jan 24 at 23:08
projectilemotion
11.5k62241
11.5k62241
asked Jan 24 at 22:21
user156720user156720
11
asked Jan 24 at 22:21
user156720user156720
11
asked Jan 24 at 22:21
user156720user156720
11
11
3
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
1
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48
add a comment |
3
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
1
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48
3
3
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
1
1
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Actually, you don't have to use double angle formulas at all...
$$tan (x) + cot 2x = 1 Rightarrow dfrac {sin x}{cos x} + dfrac {cos 2x}{sin 2x}
=1 Rightarrow dfrac {sin x sin 2x + cos x cos 2x}{cos x sin 2x} = 1$$
A couple of hints:
- What trick can you use to simplify ${sin x sin 2x + cos x cos 2x}$?
- Are there any extraneous roots? ($frac {pi}{4}$ is correct; the general solution would be $frac {pi}{4} pm pi k$.)
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
$endgroup$
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
add a comment |
$begingroup$
$tan x=t$
$$1=t+dfrac{1-t^2}{2t}$$
$$iff2t=1+t^2iff(t-1)^2=0$$
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Actually, you don't have to use double angle formulas at all...
$$tan (x) + cot 2x = 1 Rightarrow dfrac {sin x}{cos x} + dfrac {cos 2x}{sin 2x}
=1 Rightarrow dfrac {sin x sin 2x + cos x cos 2x}{cos x sin 2x} = 1$$
A couple of hints:
- What trick can you use to simplify ${sin x sin 2x + cos x cos 2x}$?
- Are there any extraneous roots? ($frac {pi}{4}$ is correct; the general solution would be $frac {pi}{4} pm pi k$.)
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
$endgroup$
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
add a comment |
$begingroup$
Actually, you don't have to use double angle formulas at all...
$$tan (x) + cot 2x = 1 Rightarrow dfrac {sin x}{cos x} + dfrac {cos 2x}{sin 2x}
=1 Rightarrow dfrac {sin x sin 2x + cos x cos 2x}{cos x sin 2x} = 1$$
A couple of hints:
- What trick can you use to simplify ${sin x sin 2x + cos x cos 2x}$?
- Are there any extraneous roots? ($frac {pi}{4}$ is correct; the general solution would be $frac {pi}{4} pm pi k$.)
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
$endgroup$
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
add a comment |
$begingroup$
Actually, you don't have to use double angle formulas at all...
$$tan (x) + cot 2x = 1 Rightarrow dfrac {sin x}{cos x} + dfrac {cos 2x}{sin 2x}
=1 Rightarrow dfrac {sin x sin 2x + cos x cos 2x}{cos x sin 2x} = 1$$
A couple of hints:
- What trick can you use to simplify ${sin x sin 2x + cos x cos 2x}$?
- Are there any extraneous roots? ($frac {pi}{4}$ is correct; the general solution would be $frac {pi}{4} pm pi k$.)
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
$endgroup$
Actually, you don't have to use double angle formulas at all...
$$tan (x) + cot 2x = 1 Rightarrow dfrac {sin x}{cos x} + dfrac {cos 2x}{sin 2x}
=1 Rightarrow dfrac {sin x sin 2x + cos x cos 2x}{cos x sin 2x} = 1$$
A couple of hints:
- What trick can you use to simplify ${sin x sin 2x + cos x cos 2x}$?
- Are there any extraneous roots? ($frac {pi}{4}$ is correct; the general solution would be $frac {pi}{4} pm pi k$.)
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
edited Jan 25 at 11:46
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
answered Jan 24 at 22:53
bjcolby15bjcolby15
1,50011016
1,50011016
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
add a comment |
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
$begingroup$
I know this solution. It 's $sin xsin2x+cos xcos 2x=cos(2x-x)=cos x$, thus we have $dfrac{cos x}{cos xsin x}=1Rightarrowdfrac{1}{sin 2x}=1Rightarrow sin2x=1$. But I think it's kind the same with double angle. Anyway thanks for answering
$endgroup$
– user156720
Jan 26 at 6:14
add a comment |
$begingroup$
$tan x=t$
$$1=t+dfrac{1-t^2}{2t}$$
$$iff2t=1+t^2iff(t-1)^2=0$$
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
$tan x=t$
$$1=t+dfrac{1-t^2}{2t}$$
$$iff2t=1+t^2iff(t-1)^2=0$$
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
add a comment |
$begingroup$
$tan x=t$
$$1=t+dfrac{1-t^2}{2t}$$
$$iff2t=1+t^2iff(t-1)^2=0$$
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
$endgroup$
$tan x=t$
$$1=t+dfrac{1-t^2}{2t}$$
$$iff2t=1+t^2iff(t-1)^2=0$$
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
answered Jan 25 at 15:35
lab bhattacharjeelab bhattacharjee
227k15158275
227k15158275
add a comment |
add a comment |
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3
$begingroup$
For what $x$? Is $x$ restricted to $[0,pi]$?
$endgroup$
– Frpzzd
Jan 24 at 22:22
1
$begingroup$
Why are you avoiding the double angle formulas? It's pretty hard to avoid them.
$endgroup$
– ConMan
Jan 24 at 22:40
$begingroup$
No Frpzzd it's not restricted.
$endgroup$
– user156720
Jan 24 at 22:42
$begingroup$
The exercise was given to me this way. I know how to solve it with double angle formulas. I don't know if it is possible to solve it without them.
$endgroup$
– user156720
Jan 24 at 22:48