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What values for theta (0<=theta<=2pi) satify the equation?

 

3 sin theta = sin theta - root3

 Jun 6, 2019
 #1
avatar+342 
+1

 

\(0≤θ≤2π \) (1)

\(3sinθ=sinθ -\sqrt{3}<=> 3sinθ-sinθ= -\sqrt{3}<=> 2sinθ= -\sqrt{3} <=>sinθ= -\frac{\sqrt{3}}{2}<=>sinθ=-sin(\frac{π}{3}),sinθ=-sin(\frac{2π}{3})<=>sinθ=sin(-\frac{π}{3}),sinθ=sin(-\frac{2π}{3})\)

With (1) \(sinθ=sin(2π-\frac{π}{3}),sinθ=sin(2π-\frac{2π}{3})<=>sinθ=sin(\frac{5π}{3}),sinθ=sin(\frac{4π}{3}) <=>θ=\frac{5π}{3},θ=\frac{4π}{3}\)

So \(θ=\frac{5π}{3},θ=\frac{4π}{3}\)

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 Jun 6, 2019
edited by Dimitristhym  Jun 6, 2019
edited by Dimitristhym  Jun 6, 2019
edited by Dimitristhym  Jun 6, 2019
edited by Dimitristhym  Jun 6, 2019
edited by Dimitristhym  Jun 6, 2019
 #2
avatar+8866 
+3

\(3\sin\theta\ =\ \sin\theta-\sqrt3\\~\\ 3\sin\theta-\sin\theta\ =\ -\sqrt3\\~\\ 2\sin\theta\ =\ -\sqrt3\\~\\ \sin\theta\ =\ -\frac{\sqrt3}{2}\)

 

We can look at a unit circle to see what values of  θ  in the interval  [0, 2π]  will satisfy  \(\sin\theta=-\frac{\sqrt3}{2}\) .

Here is a pretty clear unit circle:

https://en.wikibooks.org/wiki/Trigonometry/Trigonometric_Unit_Circle_and_Graph_Reference

 

\(\theta\ =\ \frac{4\pi}{3}\\~\\ \theta\ =\ \frac{5\pi}{3}\)

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 Jun 6, 2019
 #3
avatar+342 
0

Exactly! You have something to add in my answer or there are something whitch don't tike you?

Thank you.

Dimitristhym  Jun 6, 2019

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