help me solve this?

Question

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

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


t=_______pi

Answer 1

0 = 2*(cos t)^2 - cos t -1 = (2 * cos t + 1)*(cos t - 1). Therefore, either 2 * cos t + 1 = 0 or cos t - 1 = 0. The former requires that cos t = -1/2; on you specified interval, this occurs at t = 2*pi/3 and t = 4*pi/3. The latter requires that cos t = 1; this occurs at t = 0 and t = 2*pi.

Answer 2

cos 1pi=0,
2(cos(1pi)&^2=1,
1-cos(t)=1,
then atlast 1-1=0....
therefore the answer is 1pi,...........within the interval