I have a bunch of these problems to do but if someone can show me how to do one for example: 2((cos(t))^2)-cos(t)-1 = 0 so i have to find t with interval [0,2pi] can some one demonstrate?
I have a bunch of these problems to do but if someone can show me how to do one for example: 2((cos(t))^2)-cos(t)-1 = 0 so i have to find t with interval [0,2pi] can some one demonstrate?
Hello,Quote:
Originally Posted by mathstudent101
this equation has to be solved in 2 steps:
1. solve for cos(t)
2. solve for t
to 1.:
Substitute y = cos(t) and your equation becomes:
2y² - y - 1 = 0
which wil give: y = 1 or y = -0.5
to 2.:
Plug in the values:
cos(t) = 1 ===> t = 0 or t = 2pi
cos(t) = -0.5 ===> t = (2/3)pi or t = (4/3)pi
EB
Hello, mathstudent101!
In this problem, the angles are the "same size".
We don't need any identities.
This is a quadratic . . . factor it.Quote:
2·cos²(t) - cos(t) - 1 .= .0 . on [0, 2π]
. . [2·cos(t) + 1][cos(t) - 1] .= .0
Set each factor equal to zero and solve:
. . 2·cos(t) + 1 .= .0 . → . cos(t) = -½ . → . t = 2π/3, 4π/3
. . cos(t) - 1 .= .0 . → . cos(t) = 1 . → . t = 0