Trig question?

Question

How do I solve :2(cos(t))^2 -cos(t)-1=0

Answer 1

The trig equation 2 [cos(t)]² - cos(t) - 1 = 0

2 cos²(t) - cos(t) - 1 = 0

2 cos²(t) - 2cos(t) + cos(t) - 1 = 0

2 cos(t) [cos(t) - 1] + 1[cos(t) - 1] = 0

[cos(t) - 1] [2cos(t) + 1] = 0

 [cos(t) - 1] = 0 and 2cos(t) + 1 = 0

cos(t) = 1 and 2cos(t) = - 1

cos(t) = 1 and cos(t) = - 1/2

Solve cos(t) = 1

cos(t) = cos(0)

α = 0, t = x

The general solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

Solution t = 2nπ, where n is an integer.

 

Solve cos(t) = -1/2

Cosx is negative in second and third quadrants.

In second quadrant - π/2 ≤ x ≤ π

-1/2 = cos(π - π/3) = cos(2π/3)

The general solution of cos(x) = cos(α) is x = 2nπ ± α, where n is an integer.

Solution t = 2nπ ± 2π/3,where n is an integer.