Asked by Anonymous
Find all solutions of the equation in the interval [0,2pi)
2 cos^2 x-cos x = 0
-2cos^2 + cosx + 0
(x+1/2) (x+0/2)
(2x+1) (x+0)
-1/2,0
2Pi/3, 4pi/3, pi/2, 3pi/2
my teacher circled pi/2 and 3pi/2
What did I do wrong? I don't understand...
2 cos^2 x-cos x = 0
-2cos^2 + cosx + 0
(x+1/2) (x+0/2)
(2x+1) (x+0)
-1/2,0
2Pi/3, 4pi/3, pi/2, 3pi/2
my teacher circled pi/2 and 3pi/2
What did I do wrong? I don't understand...
Answers
Answered by
Reiny
Hard to figure out what you are doing.
2 cos^2 x-cos x = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
if cosx = 0,
x = π/2 or x = 3π/2
if cosx = 1/2
x = π/3 or x = 2π - π/3 = 5π/3
so you have 4 answers:
π/2, 3π/2, π/3, and 5π/3
(in degrees: 90°, 270°, 60°, 300° , they all work)
2 cos^2 x-cos x = 0
cosx(2cosx - 1) = 0
cosx = 0 or cosx = 1/2
if cosx = 0,
x = π/2 or x = 3π/2
if cosx = 1/2
x = π/3 or x = 2π - π/3 = 5π/3
so you have 4 answers:
π/2, 3π/2, π/3, and 5π/3
(in degrees: 90°, 270°, 60°, 300° , they all work)
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