Question
Find all solutions.In the interval (0, 2pi} of cos(2x) - 2 cos(x) = 1
Answers
Reiny
recall cos 2x = 2 cos^2 x - 1
so we have:
2cos^2 x - 1 - 2cosx = 1
2cos^2 x - 2cosx - 2 = 0
cos^2 x - cosx - 1 = 0
cosx = (1 ± √5)/2 , (ahhh thats the golden ratio)
= 1.618... , which can't happen for a cosine
or
cos x = -.61803..., angle in standard position = .90455...
x = 2.237... or x = 4.046...
so we have:
2cos^2 x - 1 - 2cosx = 1
2cos^2 x - 2cosx - 2 = 0
cos^2 x - cosx - 1 = 0
cosx = (1 ± √5)/2 , (ahhh thats the golden ratio)
= 1.618... , which can't happen for a cosine
or
cos x = -.61803..., angle in standard position = .90455...
x = 2.237... or x = 4.046...