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OpenStudy (anonymous):
Solve the following equation:
cos2x-cosx= 0; for 0≤x<2pi
Please help?
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OpenStudy (cwrw238):
first substitute for cos2x;
cos2x = 2cos^2 x - 1
to give
2cos^2 x - cos x - 1 = 0
now you need to solve this quadratic equation for cos x
OpenStudy (cwrw238):
can you do that?
OpenStudy (anonymous):
Yes! I'm doing it rn
OpenStudy (anonymous):
-1/2, right?
OpenStudy (anonymous):
I mean, there are two answers: -1/2 and 1, which one do you use? Or both?
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OpenStudy (cwrw238):
thats one solution but thers another
OpenStudy (cwrw238):
both
OpenStudy (anonymous):
That's all you have to do?
OpenStudy (cwrw238):
remember that the cosine is negative in the 2nd and 3rd segments
OpenStudy (cwrw238):
no you need values of the angles x
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OpenStudy (cwrw238):
eg one angle whose cosine is 1 is 0
OpenStudy (anonymous):
so 0º and 120º?
OpenStudy (anonymous):
120º for -1/2
OpenStudy (cwrw238):
thats 2 but there are more
for cos x = -1/2 there also a value of x in the 3rd quadrant
OpenStudy (cwrw238):
and they want solution in radians
120 degrees = 2pi/3 radians
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OpenStudy (anonymous):
How will i know which quadrant i should use though?
OpenStudy (cwrw238):
theres an easy way to remember
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