Mixing
why am I not finding pi as an extraneous solution? $2sin^2(x/2)-3sin(x/2)+1=0$
$begingroup$
$2sin^2(x/2)-3sin(x/2)+1=0$
I have solved this equation for $x = pi, pi/3, and 5pi/3$. However, after graphing this equation on desmos, I noticed that there were only 2 solutions in the range of $x = [0, 2pi)$. Why am I not getting pi as extraneous when I plug it back into the equation?
algebra-precalculus trigonometry
edited Jan 13 at 18:04
Key Flex
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
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add a comment |
$begingroup$
$2sin^2(x/2)-3sin(x/2)+1=0$
I have solved this equation for $x = pi, pi/3, and 5pi/3$. However, after graphing this equation on desmos, I noticed that there were only 2 solutions in the range of $x = [0, 2pi)$. Why am I not getting pi as extraneous when I plug it back into the equation?
algebra-precalculus trigonometry
edited Jan 13 at 18:04
Key Flex
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
$endgroup$
add a comment |
$begingroup$
$2sin^2(x/2)-3sin(x/2)+1=0$
I have solved this equation for $x = pi, pi/3, and 5pi/3$. However, after graphing this equation on desmos, I noticed that there were only 2 solutions in the range of $x = [0, 2pi)$. Why am I not getting pi as extraneous when I plug it back into the equation?
algebra-precalculus trigonometry
edited Jan 13 at 18:04
Key Flex
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
$endgroup$
$2sin^2(x/2)-3sin(x/2)+1=0$
I have solved this equation for $x = pi, pi/3, and 5pi/3$. However, after graphing this equation on desmos, I noticed that there were only 2 solutions in the range of $x = [0, 2pi)$. Why am I not getting pi as extraneous when I plug it back into the equation?
algebra-precalculus trigonometry
algebra-precalculus trigonometry
edited Jan 13 at 18:04
Key Flex
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
edited Jan 13 at 18:04
Key Flex
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
edited Jan 13 at 18:04
Key Flex
8,21761233
edited Jan 13 at 18:04
Key Flex
8,21761233
edited Jan 13 at 18:04
Key Flex
8,21761233
8,21761233
asked Jan 13 at 17:58
andrew chenandrew chen
134
asked Jan 13 at 17:58
andrew chenandrew chen
134
asked Jan 13 at 17:58
andrew chenandrew chen
134
134
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Maybe you plotted it wrong. There are three solutions in the interval $[0,2pi)$,
namely the ones you found. $pi$ is not extraneous.
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
$endgroup$
add a comment |
$begingroup$
It is $$2t^2-3t+1=0$$
so $$t^2-frac{3}{2}t+frac{1}{2}=0$$
by the quadratic formula we get
$$t_{1,2}=frac{3}{4}pmsqrt{frac{9}{16}-frac{8}{16}}$$
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
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votes
$begingroup$
Maybe you plotted it wrong. There are three solutions in the interval $[0,2pi)$,
namely the ones you found. $pi$ is not extraneous.
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
$endgroup$
add a comment |
$begingroup$
Maybe you plotted it wrong. There are three solutions in the interval $[0,2pi)$,
namely the ones you found. $pi$ is not extraneous.
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
$endgroup$
add a comment |
$begingroup$
Maybe you plotted it wrong. There are three solutions in the interval $[0,2pi)$,
namely the ones you found. $pi$ is not extraneous.
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
$endgroup$
Maybe you plotted it wrong. There are three solutions in the interval $[0,2pi)$,
namely the ones you found. $pi$ is not extraneous.
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
edited Jan 13 at 18:12
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
answered Jan 13 at 18:05
Robert IsraelRobert Israel
323k23212466
323k23212466
add a comment |
add a comment |
$begingroup$
It is $$2t^2-3t+1=0$$
so $$t^2-frac{3}{2}t+frac{1}{2}=0$$
by the quadratic formula we get
$$t_{1,2}=frac{3}{4}pmsqrt{frac{9}{16}-frac{8}{16}}$$
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
add a comment |
$begingroup$
It is $$2t^2-3t+1=0$$
so $$t^2-frac{3}{2}t+frac{1}{2}=0$$
by the quadratic formula we get
$$t_{1,2}=frac{3}{4}pmsqrt{frac{9}{16}-frac{8}{16}}$$
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
add a comment |
$begingroup$
It is $$2t^2-3t+1=0$$
so $$t^2-frac{3}{2}t+frac{1}{2}=0$$
by the quadratic formula we get
$$t_{1,2}=frac{3}{4}pmsqrt{frac{9}{16}-frac{8}{16}}$$
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
$endgroup$
It is $$2t^2-3t+1=0$$
so $$t^2-frac{3}{2}t+frac{1}{2}=0$$
by the quadratic formula we get
$$t_{1,2}=frac{3}{4}pmsqrt{frac{9}{16}-frac{8}{16}}$$
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
edited Jan 13 at 18:12
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
answered Jan 13 at 18:07
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
75.4k42865
75.4k42865
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
add a comment |
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
I think the first minus sign is a typo, isn't it?
$endgroup$
– saulspatz
Jan 13 at 18:11
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
$begingroup$
Yes, it was, thank you!
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 18:12
add a comment |
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