Solve the following equation in the interval [0, 2 pi].?

Question

Note: Give the answer as a multiple of pi. Do not use decimal numbers. The answer should be a fraction or an integer.

2(cos(t))^2-cos(t)-1=0

t=?

Answer 1

Given that

Simplify the equation

Therefore the values ot t is

 

Comments

• cos x = - 1.

cos t = cos π.

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

⇒ t = 2nπ ± π.

If, n = 0, t = 2(0)π + π and 2(0)π - π = π and - π,

If, n = 1, t = 2(1)π + π and 2(1)π - π = 3π and π.

⇒ t = π in the interval [0, 2π].

• cos t = 1/2

cos (t) = cos(π/3)

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

⇒ t = 2nπ ± (π/3)

If n = 0, t = 2(0)π + (π/3) and t = 2(0)π - (π/3) = π/3 and - π/3,

If n = 1, t = 2(1)π + (π/3) and t = 2(1)π - (π/3) = 2π + π/3 and 2π - π/3 = 7π/3 and 5π/3.

⇒ t = π/3 and t = 5π/3 in the interval [0, 2π].

Therefore, the solutions are t = π/3, t = π and t = 5π/3 in the interval [0, 2π].