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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+343 
+1

 

0θ2π (1)

3sinθ=sinθ3<=>3sinθsinθ=3<=>2sinθ=3<=>sinθ=32<=>sinθ=sin(π3),sinθ=sin(2π3)<=>sinθ=sin(π3),sinθ=sin(2π3)

With (1) sinθ=sin(2ππ3),sinθ=sin(2π2π3)<=>sinθ=sin(5π3),sinθ=sin(4π3)<=>θ=5π3,θ=4π3

So θ=5π3,θ=4π3

 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+9485 
+3

3sinθ = sinθ3 3sinθsinθ = 3 2sinθ = 3 sinθ = 32

 

We can look at a unit circle to see what values of  θ  in the interval  [0, 2π]  will satisfy  sinθ=32 .

Here is a pretty clear unit circle:

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

 

θ = 4π3 θ = 5π3

 Jun 6, 2019
 #3
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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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