Sin^2x-5/2sin2x+2=0 подробно

Решите уравнение: 2cos^2x+5sinx+1=0

2 sin^2x/2+19sinx/2-10=0 2cos^2x=cos2x+1

cos2x+1+5sinx+1=0

1-2sin^2x+1+5sinx+1=0

2sin^2x-5sinx-3=0

sinx=t

2t^2-5t-3=0

D=25+4*2*3=49

t1=-2/4=-1/2

t2 = (5+7) / 2=6 (ne mojet)

sinx=-1/2

x = (-1) ^k+1pi/6+pik sin^2x/2+19sinx/2-10=0

[2cos^2x=cos2x+1//

cos2x+1+5sinx+1=0//

1-2sin^2x+1+5sinx+1=0//

2sin^2x-5sinx-3=0

sinx=t//

2t^2-5t-3=0//

D=25+4*2*3=49//

t1=-2/4=-1/2//

t2 = (5+7) / 2=6 (ne mojet) / /

sinx=-1/2//

x = (-1) ^k/pi/6+/pi k//

2sin^2x/2+19sinx/2-10=0 / /

sinx/2=t//

2t^2+19t-10=0//

D=21^2//

t_1 = (-19-21) / 4=-10

t_2 = (-19+21) / 4=1/2

sinx/2=1/2

x/2 = (-1) ^k/pi/6+/pi*k

x = (-1) ^k/pi/3+2/pi k

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