Mixing
How many solutions has the equation $tan (3x) = 1$?
$begingroup$
I have this equation:
$tan(3x) = 1$
I've come to the following solutions:
$x_1 = 15$, so the angle is 45º
$x_2 = 75$, so the angle is 225º
But i've been told that there're up to six posible solutions fo $x$, and I don't think so. I've been watching the unit circle and there're only two possible tangents that has a value of 1.
My teacher insists that there're six solutions, but is it really possible without exceding the 360º?
trigonometry
edited Jan 24 at 15:36
Larry
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
$endgroup$
add a comment |
$begingroup$
I have this equation:
$tan(3x) = 1$
I've come to the following solutions:
$x_1 = 15$, so the angle is 45º
$x_2 = 75$, so the angle is 225º
But i've been told that there're up to six posible solutions fo $x$, and I don't think so. I've been watching the unit circle and there're only two possible tangents that has a value of 1.
My teacher insists that there're six solutions, but is it really possible without exceding the 360º?
trigonometry
edited Jan 24 at 15:36
Larry
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
$endgroup$
add a comment |
$begingroup$
I have this equation:
$tan(3x) = 1$
I've come to the following solutions:
$x_1 = 15$, so the angle is 45º
$x_2 = 75$, so the angle is 225º
But i've been told that there're up to six posible solutions fo $x$, and I don't think so. I've been watching the unit circle and there're only two possible tangents that has a value of 1.
My teacher insists that there're six solutions, but is it really possible without exceding the 360º?
trigonometry
edited Jan 24 at 15:36
Larry
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
$endgroup$
I have this equation:
$tan(3x) = 1$
I've come to the following solutions:
$x_1 = 15$, so the angle is 45º
$x_2 = 75$, so the angle is 225º
But i've been told that there're up to six posible solutions fo $x$, and I don't think so. I've been watching the unit circle and there're only two possible tangents that has a value of 1.
My teacher insists that there're six solutions, but is it really possible without exceding the 360º?
trigonometry
trigonometry
edited Jan 24 at 15:36
Larry
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
edited Jan 24 at 15:36
Larry
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
edited Jan 24 at 15:36
Larry
2,53531131
edited Jan 24 at 15:36
Larry
2,53531131
edited Jan 24 at 15:36
Larry
2,53531131
2,53531131
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
asked Jan 24 at 15:31
Slifer DragonSlifer Dragon
1084
1084
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Just $$3x=45^{circ}+180^{circ}k,$$ where $k$ is an integer number.
Now, solve the following system:
$$0^{circ}leq15^{circ}+60^{circ}kleq360^{circ}.$$
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
add a comment |
$begingroup$
Note that it isn't the "angle" (input to the tangent function) which is required to be below $360^circ$, it is $x$ itself.
So the next solution is $x_3=135^circ$. And so on.
answered Jan 24 at 15:35
ArthurArthur
118k7118201
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
$begingroup$
Just $$3x=45^{circ}+180^{circ}k,$$ where $k$ is an integer number.
Now, solve the following system:
$$0^{circ}leq15^{circ}+60^{circ}kleq360^{circ}.$$
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
add a comment |
$begingroup$
Just $$3x=45^{circ}+180^{circ}k,$$ where $k$ is an integer number.
Now, solve the following system:
$$0^{circ}leq15^{circ}+60^{circ}kleq360^{circ}.$$
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
add a comment |
$begingroup$
Just $$3x=45^{circ}+180^{circ}k,$$ where $k$ is an integer number.
Now, solve the following system:
$$0^{circ}leq15^{circ}+60^{circ}kleq360^{circ}.$$
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
$endgroup$
Just $$3x=45^{circ}+180^{circ}k,$$ where $k$ is an integer number.
Now, solve the following system:
$$0^{circ}leq15^{circ}+60^{circ}kleq360^{circ}.$$
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
edited Jan 24 at 15:38
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
answered Jan 24 at 15:34
Michael RozenbergMichael Rozenberg
108k1895200
108k1895200
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
add a comment |
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
3
3
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
$begingroup$
Sheesh I'm a few seconds late.
$endgroup$
– Don Thousand
Jan 24 at 15:34
add a comment |
$begingroup$
Note that it isn't the "angle" (input to the tangent function) which is required to be below $360^circ$, it is $x$ itself.
So the next solution is $x_3=135^circ$. And so on.
answered Jan 24 at 15:35
ArthurArthur
118k7118201
$endgroup$
add a comment |
$begingroup$
Note that it isn't the "angle" (input to the tangent function) which is required to be below $360^circ$, it is $x$ itself.
So the next solution is $x_3=135^circ$. And so on.
answered Jan 24 at 15:35
ArthurArthur
118k7118201
$endgroup$
add a comment |
$begingroup$
Note that it isn't the "angle" (input to the tangent function) which is required to be below $360^circ$, it is $x$ itself.
So the next solution is $x_3=135^circ$. And so on.
answered Jan 24 at 15:35
ArthurArthur
118k7118201
$endgroup$
Note that it isn't the "angle" (input to the tangent function) which is required to be below $360^circ$, it is $x$ itself.
So the next solution is $x_3=135^circ$. And so on.
answered Jan 24 at 15:35
ArthurArthur
118k7118201
edited Jan 24 at 15:49
answered Jan 24 at 15:35
ArthurArthur
118k7118201
answered Jan 24 at 15:35
ArthurArthur
118k7118201
answered Jan 24 at 15:35
ArthurArthur
118k7118201
118k7118201
add a comment |
add a comment |
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