2007-11-05 09:01:25 · 3 answers · asked by Adam P 1 in Science & Mathematics ➔ Mathematics
2 sin x cos x - cos x = 0
cos x (2 sin x - 1) = 0
cos x = 0 , sin x = 1 / 2
x = 90° , 270° , 30° , 150°
2007-11-05 09:21:32 · answer #1 · answered by Como 7 · 1⤊ 1⤋
Solution:
sin2x - cosx = 0
sin2x = cosx
From Double Angle Formula: sin2x = 2sinxcosx
Replacing the left side of the equation,
2sinxcosx = cosx
2sinx = cosx/cosx
2sinx = 1
sinx = 1/2
x = arcsin(1/2)
x = 30 degrees or 0.5236 radians
Hope I help.
teddyboy
2007-11-05 09:20:53 · answer #2 · answered by teddy boy 6 · 0⤊ 0⤋
cosx = sin2x = cos(pi/2 - 2x) = cos(2x - pi/2) x = pi/2 -2x + 2kpi x = pi/6 + (2/3)kpi k = 0,1,2 => x = pi/6, 5pi/6, 9pi/6 or x = 2x - pi/2 + 2kpi x = pi/2 - 2kpi k = 0 => x = pi/2
2016-05-28 00:37:32 · answer #3 · answered by marybeth 3 · 0⤊ 0⤋