Question 1203823: Find all exact solutions on [0, 2𝜋). (Enter your answers as a comma-separated list.)
2 cos^2(t) − cos(t) = 1
Found 3 solutions by MathLover1, math_tutor2020, mananth:
Answer by MathLover1(20850)   (Show Source):
Answer by math_tutor2020(3817)  (Show Source):
You can put this solution on YOUR website!
Let w = cos(t)
The original equation
2cos^2(t) - cos(t) = 1
turns into
2w^2 - w = 1
and rearranges into
2w^2 - w - 1 = 0
Use the quadratic formula, or factoring, to find the roots are:
w = -1/2 or w = 1
This leads us back to
cos(t) = -1/2 or cos(t) = 1
Then use the unit circle to find that cos(t) = -1/2 has the solutions: t = 2pi/3 and t = 4pi/3
In other words,
cos(2pi/3) = -1/2 and cos(4pi/3) = -1/2 when your calculator is set to radian mode.
Use the unit circle to find the solutions to cos(t) = 1 are: t = 0 and t = 2pi
However, we exclude 2pi due to the interval  which condenses to the interval notation [0, 2pi)
You can use a graphing calculator such as a TI83/TI84, Desmos, or GeoGebra to confirm these answers.
The function to plot would be f(x) = 2*(cos(x))^2 - cos(x) - 1
The goal is to look for the x intercepts on the interval 
Note that
2pi/3 = 2.094395 approximately
4pi/3 = 4.188790 approximately
Answers: 0, 2pi/3, 4pi/3
Answer by mananth(16946)  (Show Source):
You can put this solution on YOUR website!
2 cos^2(t) − cos(t) = 1
2 cos^2(t) − cos(t) - 1=0
2cos^2t-2cost+cost-1=0
2cost(cost -1)+1(cost-1)=0
(2cost+1)(cost-1)=0
If 2cost +1=0
2cost = -1
cost =-1/2
cost-1 = 0
cost =1
From above
This is true when
t =0,2pi/3,4pi/3
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