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anonymous
 4 years ago
Sin(2x)*Sin(x)=Cos(x) ..... Solve X
anonymous
 4 years ago
Sin(2x)*Sin(x)=Cos(x) ..... Solve X

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sin(2x)= 2Sin(x)Cos(x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So I Got ..... (2Sin(x)Cos(x)Sin(x))Cos(x)=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0try factoring them (2Sin(x)Cos(x)Sin(x))Cos(x)=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cos(x) (2 Sin(x) Sin(x))1=0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Cos(x) (2Sin(x)Sin(x))=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0hm 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@mark_o. is it possible to combine the sines?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(2Sin(x)Sin(x)) = 1 multiplying 2[Sin(x)]^2 = 1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh i didn't see that 2 was not part of sine

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok from here 2[Sin(x)]^2 = 1 [Sin(x)]^2 = 1/2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok then Sin(x)^2 = 1Cos(2x)/2 Right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nope, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How would this make it solve so i get answers to the unit circle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0([Sin(x)]^2)^1/2 = sqrt(1/2) \[\sin x=\sqrt{\frac{ 1 }{ 2 }}=\frac{ 1 }{ \sqrt{2} }*\frac{ \sqrt{2} }{ \sqrt{2} }=...?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes...you are correct x=45deg =pi/4

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0although 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Attached is a example of this..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yesssssss 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0as long as restrictions are [0,2pi)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yes you got it.. lol..:D

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh that makes so much more sense after the factoring and combining of sines... lol
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