HelpMe94
Sin(2x)*Sin(x)=Cos(x) ..... Solve X
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HelpMe94
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Sin(2x)= 2Sin(x)Cos(x)
HelpMe94
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So I Got ..... (2Sin(x)Cos(x)Sin(x))-Cos(x)=0
HelpMe94
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Now What?
mark_o.
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try factoring them
(2Sin(x)Cos(x)Sin(x))-Cos(x)=0
HelpMe94
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How?
HelpMe94
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@mark_o. How?
HelpMe94
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Cos(x) (2 Sin(x) Sin(x))-1=0
HelpMe94
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Cos(x) (2Sin(x)Sin(x))=1
mark_o.
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hm lets try this
(2Sin(x)Cos(x)Sin(x)) = Cos(x) divide both sides by cos x
then
(2Sin(x)Sin(x)) = 1
can you continue from here?
HelpMe94
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@mark_o. is it possible to combine the sines?
mark_o.
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yes
HelpMe94
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How?
mark_o.
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(2Sin(x)Sin(x)) = 1 multiplying
2[Sin(x)]^2 = 1
HelpMe94
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oh i didn't see that 2 was not part of sine
mark_o.
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ok from here
2[Sin(x)]^2 = 1
[Sin(x)]^2 = 1/2
HelpMe94
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ok then Sin(x)^2 = 1-Cos(2x)/2 Right?
mark_o.
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nope, use power of i/2 on both sides
from
[Sin(x)]^2 = 1/2
([Sin(x)]^2)^1/2 = sqrt(1/2) correct? yes or no?
HelpMe94
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How would this make it solve so i get answers to the unit circle?
mark_o.
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([Sin(x)]^2)^1/2 = sqrt(1/2)
\[\sin x=\sqrt{\frac{ 1 }{ 2 }}=\frac{ 1 }{ \sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }=...?\]
HelpMe94
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X=pi/4
mark_o.
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yes...you are correct
x=45deg =pi/4
HelpMe94
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although this seems not right there should be more to this....
because i thought....
Sin(2x)Sin(x)=Cos(x)
Sin(2x)Sin(x)-Cos(x)=0
2Sin(x)Cos(x)Sin(x)-Cos(x)=0
Cos(x)(2Sin(x)Sin(x))=1
Then Maybe.....
Cos(x)(2Sin(x)Sin(x))=1
Cos(x)=1 and 2Sin(x)=1 and Sin(x)=1
HelpMe94
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Attached is a example of this..
HelpMe94
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if it is then Cos(x)=1 has solutions 0 and 2pi
2Sin(x)=1 also as Sin(x)=1/2 solutions pi/6 and 5pi/6
Sin(x)=1 solution pi/2
mark_o.
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yesssssss thats right
2Sin(x)Cos(x)Sin(x)-Cos(x)=0
could be like
Cos(x)[(2Sin(x)Sin(x))-1]=0
cos x=0 and 2Sin(x)Sin(x))-1=0
x= arc cos 0
x=90deg=pi/2
and for
2Sin(x)Sin(x))-1]=0
2Sin(x)Sin(x))=1
HelpMe94
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as long as restrictions are [0,2pi)
mark_o.
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yes you got it.. lol..:D
HelpMe94
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oh that makes so much more sense after the factoring and combining of sines... lol