How to find $x$ such that $cos 2x - cos4x = frac12$ and $0<x<3$? [closed]
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How to solve this trigonometric equation, giving your answer in exact form:
$$cos 2x - cos4x = frac12$$ where $0 < x <3$
trigonometry
closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27
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." â amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
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How to solve this trigonometric equation, giving your answer in exact form:
$$cos 2x - cos4x = frac12$$ where $0 < x <3$
trigonometry
closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27
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." â amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
1
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00
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up vote
-1
down vote
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up vote
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down vote
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How to solve this trigonometric equation, giving your answer in exact form:
$$cos 2x - cos4x = frac12$$ where $0 < x <3$
trigonometry
How to solve this trigonometric equation, giving your answer in exact form:
$$cos 2x - cos4x = frac12$$ where $0 < x <3$
trigonometry
trigonometry
edited Sep 9 at 14:41
Bernard
112k636105
112k636105
asked Sep 9 at 13:55
Andrew Hoang
4
4
closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27
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." â amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
closed as off-topic by amWhy, José Carlos Santos, user99914, Deepesh Meena, user91500 Sep 10 at 9:27
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." â amWhy, José Carlos Santos, Community, Deepesh Meena, user91500
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
1
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00
add a comment |Â
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
1
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
1
1
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00
add a comment |Â
4 Answers
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Just for the first step.
beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign
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up vote
0
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HINT
Recall that
$$cos 4x=2cos^2 2x-1$$
then use quadratic equation.
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0
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Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
left(fracx2+fracy2right)$$
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Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$
Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
namely $$4(cos 2x)^2-2cos 2x-1=0.$$
Therefore $$cos 2x=frac1pm sqrt54,$$
which implies $$2x=arccosfrac1pm sqrt54,$$ or
$$2x=2pi-arccosfrac1pm sqrt54.$$
Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Just for the first step.
beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign
add a comment |Â
up vote
1
down vote
Just for the first step.
beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Just for the first step.
beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign
Just for the first step.
beginalign
cos2x-cos4x&=frac12 \
cos2x-(cos^22x-1)&=frac12 \
endalign
answered Sep 9 at 13:59
Yuta
63929
63929
add a comment |Â
add a comment |Â
up vote
0
down vote
HINT
Recall that
$$cos 4x=2cos^2 2x-1$$
then use quadratic equation.
add a comment |Â
up vote
0
down vote
HINT
Recall that
$$cos 4x=2cos^2 2x-1$$
then use quadratic equation.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
HINT
Recall that
$$cos 4x=2cos^2 2x-1$$
then use quadratic equation.
HINT
Recall that
$$cos 4x=2cos^2 2x-1$$
then use quadratic equation.
answered Sep 9 at 14:00
gimusi
74.4k73889
74.4k73889
add a comment |Â
add a comment |Â
up vote
0
down vote
Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
left(fracx2+fracy2right)$$
add a comment |Â
up vote
0
down vote
Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
left(fracx2+fracy2right)$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
left(fracx2+fracy2right)$$
Use that $$cos(x)-cos(y)=-2 sin left(fracx2-fracy2right) sin
left(fracx2+fracy2right)$$
answered Sep 9 at 14:01
Dr. Sonnhard Graubner
69.1k32761
69.1k32761
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add a comment |Â
up vote
0
down vote
Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$
Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
namely $$4(cos 2x)^2-2cos 2x-1=0.$$
Therefore $$cos 2x=frac1pm sqrt54,$$
which implies $$2x=arccosfrac1pm sqrt54,$$ or
$$2x=2pi-arccosfrac1pm sqrt54.$$
Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$
add a comment |Â
up vote
0
down vote
Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$
Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
namely $$4(cos 2x)^2-2cos 2x-1=0.$$
Therefore $$cos 2x=frac1pm sqrt54,$$
which implies $$2x=arccosfrac1pm sqrt54,$$ or
$$2x=2pi-arccosfrac1pm sqrt54.$$
Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$
Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
namely $$4(cos 2x)^2-2cos 2x-1=0.$$
Therefore $$cos 2x=frac1pm sqrt54,$$
which implies $$2x=arccosfrac1pm sqrt54,$$ or
$$2x=2pi-arccosfrac1pm sqrt54.$$
Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$
Notice that $$cos 4x=2cos^2 2x-1,~~~forall x in mathbbR,$$
Hence, the equation could be rewritten as $$cos 2x-(2cos^2 2x-1)=frac12,$$
namely $$4(cos 2x)^2-2cos 2x-1=0.$$
Therefore $$cos 2x=frac1pm sqrt54,$$
which implies $$2x=arccosfrac1pm sqrt54,$$ or
$$2x=2pi-arccosfrac1pm sqrt54.$$
Notice that $0<x<3$. As a result, $$x=frac12arccosfrac1pm sqrt54,$$or $$x=pi-frac12arccosfrac1pm sqrt54.$$
edited Sep 9 at 14:50
answered Sep 9 at 14:45
mengdie1982
3,774216
3,774216
add a comment |Â
add a comment |Â
You could try using the double angle formula for cos in terms of cos and solve it as a quadratic equation.
â Prashin Jeevaganth
Sep 9 at 13:58
Use the half-angle formula for $cos(4x)$ to get a quadratic equation in $cos(2x)$
â Claude Leibovici
Sep 9 at 13:58
1
We're not doing your homework for you without you showing us your efforts.
â TheSimpliFire
Sep 9 at 14:00