Asked by Sandy
Solve the following for x, where 0 < x < 2 pi.
2 cos^2 x + cos x = 1
2 cos^2 x + cos x = 1
Answers
Answered by
Damon
2 y^2 + y = 1
y^2 + (1/2)y = 1/2
y^2 + (1/2)y + 1/16 = 8/16+1/16 = 9/16
(y+1/4)^2 = 9/16
y+1/4 = -3/4 or +3/4
y = -1 or 1/2
so
cos x = -1 or cos x = 1/2
find all those places for example cos pi = -1 and cos pi/3 or 60 degrees = 1/2 as does cos of -pi/3 which is 5pi/3
y^2 + (1/2)y = 1/2
y^2 + (1/2)y + 1/16 = 8/16+1/16 = 9/16
(y+1/4)^2 = 9/16
y+1/4 = -3/4 or +3/4
y = -1 or 1/2
so
cos x = -1 or cos x = 1/2
find all those places for example cos pi = -1 and cos pi/3 or 60 degrees = 1/2 as does cos of -pi/3 which is 5pi/3
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