ответ:
x=+-п/6+пk
объяснение:
8cos2x=2cos4x+5
8cos2x=2(2cos^22x-1)+5
8cos2x=4cos^2(2x)+3
cos2x=t |t|< =1
8t=4t^2+3
4t^2-8t+3=0
t=(4+-sqrt(16-12))/4
t=(4+-2)/4
t=3/2 t=1/2
cos2x=1/2
x=+-п/6+пk
d = b^2 - 4ac = (-9)^2 - 4·5·(-2) = 81 + 40 = 121
x1 = (9 - √121)/2·5 = (9 - 11)/10 = -2/10 = -0.2
x2 = (9 + √121)/2·5 = (9 + 11)/10 = 20/10 = 2
б)2x^2 + 3x - 2 = 0
d = b^2 - 4ac = 3^2 - 4·2·(-2) = 9 + 16 = 25
x1 = ( -3 - √25)/2·2 = ( -3 - 5)/4 = -8/4 = -2x2 = (-3 + √25)/2·2 = (-3 + 5)/4 = 2/4 = 0.5
в)2x^2 + 7x + 3 = 0
d = b^2 - 4ac = 7^2 - 4·2·3 = 49 - 24 = 25
x1 = (-7 - √25)/2·2 = (-7 - 5)/4 = -12/4 = -3
x2 = (-7 + √25)/2·2 = (-7 + 5)/4 = -2/4 = -0.5
г)5x^2 - 8x - 4 = 0
d = b^2 - 4ac = (-8)^2 - 4·5·(-4) = 64 + 80 = 144
x1 = ( 8 - √144)/2·5 = ( 8 - 12)/10 = -4/10 = -0.4x2 = (8 + √144)/2·5 = (8 + 12)/10 = 20/10 = 2
Поделитесь своими знаниями, ответьте на вопрос:
5tgx-8ctgx+6=0 |*tgx
5tg^2x+6tgx-8=0
tgx=t
5t^2+6t-8=0
d=36-4*5*(-8)=196
t=-2
t=4/5
1)tgx=-2
x=arctg(-2)+pik . k=z
2)tgx=4/5
x=arctg(4/5)+pik . k=z
------------------------------------------------
sin2x+1=4cos^2x
2sinxcosx=4cos^2x-1 | /cos^2x
2tgx=4-1/cos^2x
-2tgx=(1/cos^2x -1) -3
-2tgx=tg^2x-3
tg^2x+2tgx-3=0
tgx=t
t^2+2t-3=0
t=1
t=-3
1)tgx=1
x=pi/4+pik . k=z
2)tgx=-3
x=arctg(-3)+pik . k=z
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14cos^2x+3=3cos^2x-10sin^2x
11cos^2x+10sin^2x+3=0
10+cos^2x+3=0
cosx^2=-13 -нет корней