Solving $sec^2(x)-2sqrt{2} sec(x)+2=0$ [closed]
What are the values of $x$ in degrees when:
$$
sec^2(x)-2sqrt{2} sec(x)+2=0
$$
I believe I should use derivatives of $sec(x)$ but am struggling.
Any help would be appreciated.
trigonometry
edited Dec 3 '18 at 14:26
Blue
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
closed as off-topic by Saad, Alexander Gruber♦ Dec 4 '18 at 4:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
What are the values of $x$ in degrees when:
$$
sec^2(x)-2sqrt{2} sec(x)+2=0
$$
I believe I should use derivatives of $sec(x)$ but am struggling.
Any help would be appreciated.
trigonometry
edited Dec 3 '18 at 14:26
Blue
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
closed as off-topic by Saad, Alexander Gruber♦ Dec 4 '18 at 4:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27
add a comment |
What are the values of $x$ in degrees when:
$$
sec^2(x)-2sqrt{2} sec(x)+2=0
$$
I believe I should use derivatives of $sec(x)$ but am struggling.
Any help would be appreciated.
trigonometry
edited Dec 3 '18 at 14:26
Blue
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
What are the values of $x$ in degrees when:
$$
sec^2(x)-2sqrt{2} sec(x)+2=0
$$
I believe I should use derivatives of $sec(x)$ but am struggling.
Any help would be appreciated.
trigonometry
trigonometry
edited Dec 3 '18 at 14:26
Blue
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
edited Dec 3 '18 at 14:26
Blue
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
edited Dec 3 '18 at 14:26
Blue
47.7k870151
edited Dec 3 '18 at 14:26
Blue
47.7k870151
edited Dec 3 '18 at 14:26
Blue
47.7k870151
47.7k870151
asked Dec 3 '18 at 13:25
user8469209user8469209
82
asked Dec 3 '18 at 13:25
user8469209user8469209
82
asked Dec 3 '18 at 13:25
user8469209user8469209
82
82
closed as off-topic by Saad, Alexander Gruber♦ Dec 4 '18 at 4:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Alexander Gruber♦ Dec 4 '18 at 4:21
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Alexander Gruber
If this question can be reworded to fit the rules in the help center, please edit the question.
1
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27
add a comment |
1
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27
1
1
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27
add a comment |
2 Answers
2
active
oldest
votes
Hint:
the equation is:
$$
left(sec x -sqrt{2}right)^2=0
$$
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
add a comment |
Hint:
This is a quadratic, so you can use the substitution $u = sec x$, which yields
$$u^2-2sqrt2u+2 = 0$$
which so happens to be a perfect square trinomial which can be factored, yielding
$$(u-sqrt 2)^2 = 0$$
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint:
the equation is:
$$
left(sec x -sqrt{2}right)^2=0
$$
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
add a comment |
Hint:
the equation is:
$$
left(sec x -sqrt{2}right)^2=0
$$
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
add a comment |
Hint:
the equation is:
$$
left(sec x -sqrt{2}right)^2=0
$$
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
Hint:
the equation is:
$$
left(sec x -sqrt{2}right)^2=0
$$
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
answered Dec 3 '18 at 13:30
Emilio NovatiEmilio Novati
51.6k43473
51.6k43473
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
add a comment |
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
Therefore sec x =√2 and is +/- 45 degress
– user8469209
Dec 3 '18 at 13:44
add a comment |
Hint:
This is a quadratic, so you can use the substitution $u = sec x$, which yields
$$u^2-2sqrt2u+2 = 0$$
which so happens to be a perfect square trinomial which can be factored, yielding
$$(u-sqrt 2)^2 = 0$$
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
add a comment |
Hint:
This is a quadratic, so you can use the substitution $u = sec x$, which yields
$$u^2-2sqrt2u+2 = 0$$
which so happens to be a perfect square trinomial which can be factored, yielding
$$(u-sqrt 2)^2 = 0$$
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
add a comment |
Hint:
This is a quadratic, so you can use the substitution $u = sec x$, which yields
$$u^2-2sqrt2u+2 = 0$$
which so happens to be a perfect square trinomial which can be factored, yielding
$$(u-sqrt 2)^2 = 0$$
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
Hint:
This is a quadratic, so you can use the substitution $u = sec x$, which yields
$$u^2-2sqrt2u+2 = 0$$
which so happens to be a perfect square trinomial which can be factored, yielding
$$(u-sqrt 2)^2 = 0$$
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
answered Dec 3 '18 at 13:27
KM101KM101
5,5511423
5,5511423
add a comment |
add a comment |
1
I believe that before trying to find out what $x$ is, try to find out what $sec(x)$ might be.
– Arthur
Dec 3 '18 at 13:27