Interval in which $(cos p-1)x^2+cos p.x+sin p=0$ has real roots
The equation $(cos p-1)x^2+cos p.x+sin p=0$ where $x$ is a variable has real roots. Then the interval of $p$ may be any of the following:
$$
(a)quad(0,2pi)quad (b)quad (-pi,0)quad (c)quad big(frac{-pi}{2},frac{pi}{2}big)quad (d)quad(0,pi)
$$
The solution given in my reference is the interval $(0,pi)$.
My Attempt
$$
Delta=cos^2p-4(cos p-1)sin pgeq 0\
cos^2p-4sin pcos p+4sin pgeq 0\
Delta'=16sin^2p-16sin pleq0impliessin^2pleqsin p\
$$
Similar problem has been asked before Find the range of values of $p$ if $(cos p−1)x^2+(cos p)x+sin p=0$ has real roots in the variable $x$. but it does not address how to prove it analytically.
trigonometry
asked Nov 29 '18 at 19:10
ss1729
1,8491723
add a comment |
The equation $(cos p-1)x^2+cos p.x+sin p=0$ where $x$ is a variable has real roots. Then the interval of $p$ may be any of the following:
$$
(a)quad(0,2pi)quad (b)quad (-pi,0)quad (c)quad big(frac{-pi}{2},frac{pi}{2}big)quad (d)quad(0,pi)
$$
The solution given in my reference is the interval $(0,pi)$.
My Attempt
$$
Delta=cos^2p-4(cos p-1)sin pgeq 0\
cos^2p-4sin pcos p+4sin pgeq 0\
Delta'=16sin^2p-16sin pleq0impliessin^2pleqsin p\
$$
Similar problem has been asked before Find the range of values of $p$ if $(cos p−1)x^2+(cos p)x+sin p=0$ has real roots in the variable $x$. but it does not address how to prove it analytically.
trigonometry
asked Nov 29 '18 at 19:10
ss1729
1,8491723
add a comment |
The equation $(cos p-1)x^2+cos p.x+sin p=0$ where $x$ is a variable has real roots. Then the interval of $p$ may be any of the following:
$$
(a)quad(0,2pi)quad (b)quad (-pi,0)quad (c)quad big(frac{-pi}{2},frac{pi}{2}big)quad (d)quad(0,pi)
$$
The solution given in my reference is the interval $(0,pi)$.
My Attempt
$$
Delta=cos^2p-4(cos p-1)sin pgeq 0\
cos^2p-4sin pcos p+4sin pgeq 0\
Delta'=16sin^2p-16sin pleq0impliessin^2pleqsin p\
$$
Similar problem has been asked before Find the range of values of $p$ if $(cos p−1)x^2+(cos p)x+sin p=0$ has real roots in the variable $x$. but it does not address how to prove it analytically.
trigonometry
asked Nov 29 '18 at 19:10
ss1729
1,8491723
The equation $(cos p-1)x^2+cos p.x+sin p=0$ where $x$ is a variable has real roots. Then the interval of $p$ may be any of the following:
$$
(a)quad(0,2pi)quad (b)quad (-pi,0)quad (c)quad big(frac{-pi}{2},frac{pi}{2}big)quad (d)quad(0,pi)
$$
The solution given in my reference is the interval $(0,pi)$.
My Attempt
$$
Delta=cos^2p-4(cos p-1)sin pgeq 0\
cos^2p-4sin pcos p+4sin pgeq 0\
Delta'=16sin^2p-16sin pleq0impliessin^2pleqsin p\
$$
Similar problem has been asked before Find the range of values of $p$ if $(cos p−1)x^2+(cos p)x+sin p=0$ has real roots in the variable $x$. but it does not address how to prove it analytically.
trigonometry
trigonometry
asked Nov 29 '18 at 19:10
ss1729
1,8491723
asked Nov 29 '18 at 19:10
ss1729
1,8491723
asked Nov 29 '18 at 19:10
ss1729
1,8491723
asked Nov 29 '18 at 19:10
ss1729
1,8491723
asked Nov 29 '18 at 19:10
ss1729
1,8491723
1,8491723
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Adding to what you did, Let us solve the inequation
$$sin^2 ple sin p$$
or
$$sin p(sin p-1)le 0$$
which yields to
$$sin pge 0$$ because $; (sin p-1)le 0$.
the answer is $d) : 0<p<pi$.
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
add a comment |
Hint :
$$cos^2p - 4sin p cos p + 4sin p = (cos p - sin p)^2 +4sin p - sin^2p$$
$$=$$
$$(cos p - sin p )^2 + (4-sin p)sin p $$
Thus, the consraint yielded for real solutions is translated to :
$$(cos p - sin p )^2 + (4-sin p)sin p geq 0$$
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
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
Adding to what you did, Let us solve the inequation
$$sin^2 ple sin p$$
or
$$sin p(sin p-1)le 0$$
which yields to
$$sin pge 0$$ because $; (sin p-1)le 0$.
the answer is $d) : 0<p<pi$.
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
add a comment |
Adding to what you did, Let us solve the inequation
$$sin^2 ple sin p$$
or
$$sin p(sin p-1)le 0$$
which yields to
$$sin pge 0$$ because $; (sin p-1)le 0$.
the answer is $d) : 0<p<pi$.
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
add a comment |
Adding to what you did, Let us solve the inequation
$$sin^2 ple sin p$$
or
$$sin p(sin p-1)le 0$$
which yields to
$$sin pge 0$$ because $; (sin p-1)le 0$.
the answer is $d) : 0<p<pi$.
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
Adding to what you did, Let us solve the inequation
$$sin^2 ple sin p$$
or
$$sin p(sin p-1)le 0$$
which yields to
$$sin pge 0$$ because $; (sin p-1)le 0$.
the answer is $d) : 0<p<pi$.
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
answered Nov 29 '18 at 19:49
hamam_Abdallah
38k21634
38k21634
add a comment |
add a comment |
Hint :
$$cos^2p - 4sin p cos p + 4sin p = (cos p - sin p)^2 +4sin p - sin^2p$$
$$=$$
$$(cos p - sin p )^2 + (4-sin p)sin p $$
Thus, the consraint yielded for real solutions is translated to :
$$(cos p - sin p )^2 + (4-sin p)sin p geq 0$$
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
add a comment |
Hint :
$$cos^2p - 4sin p cos p + 4sin p = (cos p - sin p)^2 +4sin p - sin^2p$$
$$=$$
$$(cos p - sin p )^2 + (4-sin p)sin p $$
Thus, the consraint yielded for real solutions is translated to :
$$(cos p - sin p )^2 + (4-sin p)sin p geq 0$$
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
add a comment |
Hint :
$$cos^2p - 4sin p cos p + 4sin p = (cos p - sin p)^2 +4sin p - sin^2p$$
$$=$$
$$(cos p - sin p )^2 + (4-sin p)sin p $$
Thus, the consraint yielded for real solutions is translated to :
$$(cos p - sin p )^2 + (4-sin p)sin p geq 0$$
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
Hint :
$$cos^2p - 4sin p cos p + 4sin p = (cos p - sin p)^2 +4sin p - sin^2p$$
$$=$$
$$(cos p - sin p )^2 + (4-sin p)sin p $$
Thus, the consraint yielded for real solutions is translated to :
$$(cos p - sin p )^2 + (4-sin p)sin p geq 0$$
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
answered Nov 29 '18 at 19:27
Rebellos
14.5k31245
14.5k31245
add a comment |
add a comment |
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