Question 1203823
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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: <font color=red>t = 2pi/3 and t = 4pi/3</font>
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: <font color=red>t = 0</font> and t = 2pi
However, we exclude 2pi due to the interval {{{0 <= t < 2pi}}} 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 {{{0 <= x < 2pi}}}
Note that
2pi/3 = 2.094395 approximately
4pi/3 = 4.188790 approximately




Answers:  <font color=red size=4>0, 2pi/3, 4pi/3</font>
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