Cos3x*cosx=cos2x

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Axdefb57853

17 апр. 2014 г., 0:52:13 (10 лет назад)

cos3x*cosx = cos2x
1/2*[cos(3x-x) + cos(3x + x)] = cos2x
cos2x - cos4x = 2 cos2x
cos4x - cos2x = 0
2*[sin(4x + 2x)/2 * sin(2x - 4x)/2] = 0
1)  sin3x = 0
3x = πn, n∈Z
x = πn/3, n∈z
2)  sinx = 0
x =  πk, k∈Z
 

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Ninelzol

17 апр. 2014 г., 2:29:24 (10 лет назад)

1/2cos2x+1/2cos4x-cos2x=0
1/2cos4x-1/2cos2x=0
1/2*(2cos²2x-1)-1/2cos2x=0
cos²2x-1/2cos2x-1/2=0
cos2x=a
a²-0,5a-0,5a=0
D=0,25+2=2,25
a1=(0,5-1,5)/2=-0,5⇒cos2x=-0,5⇒2x=+-5π/6+2πn⇒x=+-5π/12+πn
a2=(0,5+1,5)/2=1⇒cos2x=1⇒2x=2πn⇒x=πn