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Sin(2x)*Sin(x)=Cos(x) ..... Solve X

Mathematics
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Sin(2x)= 2Sin(x)Cos(x)
So I Got ..... (2Sin(x)Cos(x)Sin(x))-Cos(x)=0
Now What?

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Other answers:

try factoring them (2Sin(x)Cos(x)Sin(x))-Cos(x)=0
How?
@mark_o. How?
Cos(x) (2 Sin(x) Sin(x))-1=0
Cos(x) (2Sin(x)Sin(x))=1
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?
@mark_o. is it possible to combine the sines?
yes
How?
(2Sin(x)Sin(x)) = 1 multiplying 2[Sin(x)]^2 = 1
oh i didn't see that 2 was not part of sine
ok from here 2[Sin(x)]^2 = 1 [Sin(x)]^2 = 1/2
ok then Sin(x)^2 = 1-Cos(2x)/2 Right?
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?
How would this make it solve so i get answers to the unit circle?
([Sin(x)]^2)^1/2 = sqrt(1/2) \[\sin x=\sqrt{\frac{ 1 }{ 2 }}=\frac{ 1 }{ \sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }=...?\]
X=pi/4
yes...you are correct x=45deg =pi/4
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
Attached is a example of this..
1 Attachment
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
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
as long as restrictions are [0,2pi)
yes you got it.. lol..:D
oh that makes so much more sense after the factoring and combining of sines... lol

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