SOLUTION: Solve exactly, where possible, giving answers in radians. Approximate answers must be rounded to 2 decimal places:
12cos^2 x-4cosx =1 0≤x<2π
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-> SOLUTION: Solve exactly, where possible, giving answers in radians. Approximate answers must be rounded to 2 decimal places:
12cos^2 x-4cosx =1 0≤x<2π
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Question 1161422: Solve exactly, where possible, giving answers in radians. Approximate answers must be rounded to 2 decimal places:
12cos^2 x-4cosx =1 0≤x<2π Found 2 solutions by Edwin McCravy, MathLover1:Answer by Edwin McCravy(20066) (Show Source):
Factor the left side:
6cos(x)+1=0; 2cos(x)-1=0
6cos(x)=-1; 2cos(x)=1
cos(x)=-1/6; cos(x)=1/2
For cos(x)=-1/6 we know that x is in QII or QIII. First we find the angle in
QI that has +1/6 for its cosine. [It will not be a solution but we can get
the others from that reference angle.
We find the inverse cosine of +1/6 to be 1.403348238. That's the reference
angle. To get the QII answer we subtract from π
π-1.403348238 = 1.738244406, rounds to 1.74 <--QII answer
To get the QIII answer we add to π
π+1.403348238 = 4.544940902, rounds to 4.54 <--QIII answer
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For cos(x)=1/2 we know that x is in QI or QIV. Those are special angles
that we get off the unit circle.
They are π/3 and 5π/3
So there are 4 answers, two of them exact, and the other two approximated.
Edwin
