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sq3tanx=2sinx I am having trouble breaking these down so if you could answer with the steps that would be great :) 0<=x<2pi
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okay put sin(x) and tan(x) in one side and the other in the other side
tan(x)/sin(x)=2/sq(3)
tan(x)=sin(x)/cos(x) so....(sin(x)/cos(x))/sin(x))=sin(x)/(cos(x)*sin(x)=1/cos(x)
Our initial equation:\[\sqrt{3}\tan(x)=2sinx\] Express tan(x) in terms of sin(x) and cos(x) \[\sqrt{3}\frac{ \sin(x) }{ \cos(x) } = 2\sin(x)\] Divide by sin(x) (Note: this is potentially 0 so we must remember a solution is sin(x)=0 later on!) \[\sqrt{3}\frac{ 1 }{ \cos(x) } = 2\] Rearrange to get an equation for cos(x): \[\cos(x)=\frac{ \sqrt{3} }{ 2 }\] Our solutions are: x=0, x=pi/6, x=11pi/6
so cos(x)=sq(3)/2
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I thank all of you :)
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