What values for theta (0<=theta<=2pi) satify the equation?
3 sin theta = sin theta - root3
0≤θ≤2π (1)
3sinθ=sinθ−√3<=>3sinθ−sinθ=−√3<=>2sinθ=−√3<=>sinθ=−√32<=>sinθ=−sin(π3),sinθ=−sin(2π3)<=>sinθ=sin(−π3),sinθ=sin(−2π3)
With (1) sinθ=sin(2π−π3),sinθ=sin(2π−2π3)<=>sinθ=sin(5π3),sinθ=sin(4π3)<=>θ=5π3,θ=4π3
So θ=5π3,θ=4π3
3sinθ = sinθ−√3 3sinθ−sinθ = −√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
Exactly! You have something to add in my answer or there are something whitch don't tike you?
Thank you.