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?

- Feb 15th 2007, 09:38 PMmathstudent101Solving a Multiple angle equation using double angle formulas
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?

- Feb 15th 2007, 11:00 PMearboth
Hello,

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 - Feb 15th 2007, 11:16 PMSoroban
Hello, mathstudent101!

In this problem, the angles are the "same size".

We don't need any identities.

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