Cos2x Cosx 0
w sol2 solves cosx = 0? ۔
cos (2x) cos (x) = 0.
=> 2cos 2 (x) 1 cos x = 0.
=> 2cos 2 (x) 2cos x + cos x 1 = 0.
=> (2 cos x + 1) (cos x 1) = 0.
Case 1: 2cos x + 1 = 0.
cos x = 1/2.
x = 2ÃÂ / 3, 4ÃÂ / 3, (2ÃÂ / 3 + 2nÃÂ) and (4ÃÂ / 3 + 2nÃÂ)
Case 2: cos x 1 = 0.
cos = 1.
x = 0.2no.
x = 0, 2ÃÂ / 3, 4Ã / 3, (2Â / 3 + 2nÃŽÂ), 2nÃÂ and (4Â / 3 + 2nÃŽÂ) where n is a number
Cos2x Cosx 0
Suppose cos2x is cos 2x, not cos² x. I also used the degree of 0 x 360 in degrees.
Extend cos 2x with the cosine double angle formula, ie:
cos 2x cos' cos x = 0
X 2 cos² x ˆ '1 ÂÂ' cos x = 0
Let y = cos x
² 2 years ² 'and ˆÂ' 1 = 0.
ˆ´ (y ˆ '1) (2y + 1) = 0.
ˆ´ y = ˆ´Ã½ or 1.
ˆ´ cos x = ˆÂ'½ o 1
ˆ´ x = arccos (ȴý) or arccos (1)
´ x = 120, 360 ˆ 120, 0, 360 ˆÂ'0.
´ x = 0 °, 120Â, 240Â, 0, 360Â.
NS
Cos2x Cosx 0
Cos2x Cosx 0
w sol2 solves cosx = 0? 3
cos (2x) cos (x) = 0
=> 2cos 2 (x) 1 cos x = 0
=> 2cos 2 (x) 2cos x + cos x 1 = 0
=> (2 cos x + 1) (cos x 1) = 0
Case 1: 2cos x + 1 = 0
cos x = 1/2
x = 2ÃÂ / 3, 4ÃÂ / 3, (2ÃÂ / 3 + 2nÃÂ) and (4ÃÂ / 3 + 2nÃÂ)
Case 2: cos x 1 = 0
cos = 1
x = 0.2 number
x = 0, 2ÃÂ / 3, 4Ã / 3, (2Â / 3 + 2nÃŽÂ), 2nÃÂ and (4Â / 3 + 2nÃŽÂ) where n is a number
Cos2x Cosx 0
Cos2x Cosx 0
Suppose cos2x is cos 2x, not cos² x. I also used the degree of 0 x 360 in degrees.
Extend cos 2x with the cosine double angle formula, ie:
cos 2x  'cos x = 0
2 cos² x ˆ '1 ÂÂ' cos x = 0
Let y = cos x
 '2 years' and' 1 = 0
 '(y Â'1) (2y + 1) = 0
´y = ˆ´Ã½ or 1
Cos' cos x = Â'½ o 1
´ x = arccos (ôý) or arccos (1)
´ x = 120, 360 120, 0, 360Â'0
´ x = 0 °, 120 °, 240 °, 0, 360Â
NS
cos2x cosx = 0
2 sins x cos x cos x = 0
cos x (2 sin x 1) = 0
cos x = 0: x = pi / 2 and x = 3 pi / 2
2 sins x 1 = 0
Sin x = 1/2: x = pi / 6 and x = 5 pi / 6
Use the double angle formula and factor cos2x = 2cos 2 x 1 cos x (2 cos 2 x 1) = 0 then cos x = 0 and cos x = + 1 / square 2
Is it cos (2x) or cos (x) 2?
However, here it is:
cos (2x) cos (x) = 0
cos (x) 2 sin (x) 2 cos (x) = 0
cos (x) 2 (1 cos (x) 2) cos (x) = 0
cos (x) 2 1 + cos (x) 2 cos (x) = 0
2cos (x) 2cos (x) 1 = 0
cos (x) = (1 + / square (1 + 4 * 2)) / 4
cos (x) = (1 + / square (9)) / 4
cos (x) = (1 + / 3) / 4
cos (x) = 1, 1/2
x = 2pi * k, 2pi / 3 + 2pi * k, 4pi / 3 + 2pi * k: k is a number
cos (x) 2 cos (x) = 0
cos (x) * (cos (x) 1) = 0
cos (x) = 0
x = pi / 2 + pi * k
cos (x) = 1
x = 2 * pi * k
k is a number.
