(radical 3)(sin t) + cos t = 1 find solutions between 0 and 2 pi. How would I do this? I was thinking that I could move the cos t to the other side then square everything. Then I could change the right side to sin^2t and then move it back over and factor. Could someone show me how you would do it?
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√3 sin(t) + cos(t) = 1
√3 sin(t) = 1 - cos(t)
3 sin²(t) = (1 - cos(t))²
3 (1 - cos²(t)) = 1 - 2 cos(t) + cos²(t)
3 - 3 cos²(t) = 1 - 2 cos(t) + cos²(t)
0 = 4 cos²(t) - 2 cos(t) - 2
0 = 2 (2 cos²(t) - cos(t) - 1)
0 = (2 cos(t) + 1) (cos(t) - 1)
Now you have the equations
(i) 2 cos(t) + 1 = 0
(ii) cos(t) - 1 = 0
(i) 2 cos(t) = -1
cos(t) = -1/2
t = 2π/3 and 4π/3
(ii) cos(t) = 1
t = 0 and 2π
(depending on the interval the question gives you)
√3 sin(t) = 1 - cos(t)
3 sin²(t) = (1 - cos(t))²
3 (1 - cos²(t)) = 1 - 2 cos(t) + cos²(t)
3 - 3 cos²(t) = 1 - 2 cos(t) + cos²(t)
0 = 4 cos²(t) - 2 cos(t) - 2
0 = 2 (2 cos²(t) - cos(t) - 1)
0 = (2 cos(t) + 1) (cos(t) - 1)
Now you have the equations
(i) 2 cos(t) + 1 = 0
(ii) cos(t) - 1 = 0
(i) 2 cos(t) = -1
cos(t) = -1/2
t = 2π/3 and 4π/3
(ii) cos(t) = 1
t = 0 and 2π
(depending on the interval the question gives you)