Number of solutions to a given equation
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Any hints on how to proceed further?
trigonometry
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
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add a comment |
$begingroup$
Any hints on how to proceed further?
trigonometry
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
$endgroup$
$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
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– Mindlack
Jan 21 at 17:19
add a comment |
$begingroup$
Any hints on how to proceed further?
trigonometry
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
$endgroup$
Any hints on how to proceed further?
trigonometry
trigonometry
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
asked Jan 21 at 17:16
Sameer ThakurSameer Thakur
305
305
$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
$endgroup$
– Mindlack
Jan 21 at 17:19
add a comment |
$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
$endgroup$
– Mindlack
Jan 21 at 17:19
$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
$endgroup$
– Mindlack
Jan 21 at 17:19
$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
$endgroup$
– Mindlack
Jan 21 at 17:19
add a comment |
2 Answers
2
active
oldest
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$begingroup$
Put $;t=sin x;$, then the equation becomes
$$2^{t^2}+5cdot 2^{1-t^2}=7iff2^{2t^2}-7cdot 2^{t^2}+10=0$$
Put now $;y:=2^{t^2};$ and get the quadratic
$$y^2-7y+10=0implies (y-5)(y-2)=0impliesbegin{cases}5=y_1=2^{t^2}implies t^2=frac{log5}{log2}...etc.\{}\
2=y_2=2^{t^2}implies t^2=1...etc.end{cases}$$
Observe that $;t=sin x;$ and thus only one of the above two options for $;t^2;$ is possible...end now the exercise.
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
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add a comment |
$begingroup$
Hint: try with $$sin^2x+cos^2x =1$$
Say $t= sin^2x $ and then $cos^2x =1-t$...
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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active
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active
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votes
$begingroup$
Put $;t=sin x;$, then the equation becomes
$$2^{t^2}+5cdot 2^{1-t^2}=7iff2^{2t^2}-7cdot 2^{t^2}+10=0$$
Put now $;y:=2^{t^2};$ and get the quadratic
$$y^2-7y+10=0implies (y-5)(y-2)=0impliesbegin{cases}5=y_1=2^{t^2}implies t^2=frac{log5}{log2}...etc.\{}\
2=y_2=2^{t^2}implies t^2=1...etc.end{cases}$$
Observe that $;t=sin x;$ and thus only one of the above two options for $;t^2;$ is possible...end now the exercise.
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Put $;t=sin x;$, then the equation becomes
$$2^{t^2}+5cdot 2^{1-t^2}=7iff2^{2t^2}-7cdot 2^{t^2}+10=0$$
Put now $;y:=2^{t^2};$ and get the quadratic
$$y^2-7y+10=0implies (y-5)(y-2)=0impliesbegin{cases}5=y_1=2^{t^2}implies t^2=frac{log5}{log2}...etc.\{}\
2=y_2=2^{t^2}implies t^2=1...etc.end{cases}$$
Observe that $;t=sin x;$ and thus only one of the above two options for $;t^2;$ is possible...end now the exercise.
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
$endgroup$
add a comment |
$begingroup$
Put $;t=sin x;$, then the equation becomes
$$2^{t^2}+5cdot 2^{1-t^2}=7iff2^{2t^2}-7cdot 2^{t^2}+10=0$$
Put now $;y:=2^{t^2};$ and get the quadratic
$$y^2-7y+10=0implies (y-5)(y-2)=0impliesbegin{cases}5=y_1=2^{t^2}implies t^2=frac{log5}{log2}...etc.\{}\
2=y_2=2^{t^2}implies t^2=1...etc.end{cases}$$
Observe that $;t=sin x;$ and thus only one of the above two options for $;t^2;$ is possible...end now the exercise.
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
$endgroup$
Put $;t=sin x;$, then the equation becomes
$$2^{t^2}+5cdot 2^{1-t^2}=7iff2^{2t^2}-7cdot 2^{t^2}+10=0$$
Put now $;y:=2^{t^2};$ and get the quadratic
$$y^2-7y+10=0implies (y-5)(y-2)=0impliesbegin{cases}5=y_1=2^{t^2}implies t^2=frac{log5}{log2}...etc.\{}\
2=y_2=2^{t^2}implies t^2=1...etc.end{cases}$$
Observe that $;t=sin x;$ and thus only one of the above two options for $;t^2;$ is possible...end now the exercise.
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
answered Jan 21 at 17:33
DonAntonioDonAntonio
179k1494230
179k1494230
add a comment |
add a comment |
$begingroup$
Hint: try with $$sin^2x+cos^2x =1$$
Say $t= sin^2x $ and then $cos^2x =1-t$...
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
$endgroup$
add a comment |
$begingroup$
Hint: try with $$sin^2x+cos^2x =1$$
Say $t= sin^2x $ and then $cos^2x =1-t$...
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
$endgroup$
add a comment |
$begingroup$
Hint: try with $$sin^2x+cos^2x =1$$
Say $t= sin^2x $ and then $cos^2x =1-t$...
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
$endgroup$
Hint: try with $$sin^2x+cos^2x =1$$
Say $t= sin^2x $ and then $cos^2x =1-t$...
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
answered Jan 21 at 17:18
greedoidgreedoid
44.2k1155110
44.2k1155110
add a comment |
add a comment |
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$begingroup$
Write $y=2^{(sin{x})^2}$, you want the number of solutions to $y(x)+10/y(x)=7$. So I suggest that you find the solutions to $y+10/y=7$ and then find out how many $x$ give those values.
$endgroup$
– Mindlack
Jan 21 at 17:19