Знатоки, прошу вашей помощи. а) Ctg^2(x/3+п/4) = 3;

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AlenaAkhmetshina

22 мая 2013 г., 14:58:43 (10 лет назад)

а) Ctg^2(x/3+п/4) = 3;
1+ctg²(x/3+π/4)=4
1/sIn²(x/3+π/4)=4
sin²(x/3+π/4)=1/4
(1-cos(2x/3+π/2)/2=1/4
1+sin(2x/3)=1/2
sin(2x/3)=-1/2
2x/3=(-1)^n+1 * π/6+πn
x=(-1)^n+1 * π/4+3πn/2
б) Tg^2+tgx =2;
tgx=a
a²+a-2=0
a1+a2=-1 U a1*a2=-2
a1=-2⇒tgx=-2⇒x=-arctg2+πn
a2=1⇒tgx=1⇒x=π/4
в) 2cos^2x + 3 sinx = 0;
2-2sin²x+3sinx=0
sinx=a
2a²-3a-2=0
D=9+16=25
a1=(3-5)/4=-1/2⇒sinx=-1/2⇒x=(-1)^n+1 *π/6+πn
a2=(3+5)/4=2⇒sinx=2∉[-1;1]
г) 7sin(x/7 - п/3) = 0;
sin(x/7 - п/3) = 0;
x/7-π/3=πn
x/7=π/3 +πn
x=7π/3+7πn
д) 2cos (x/2 + п/3) = -1
cos (x/2 + п/3) = -1/2
x/2+π/3=2π/3+2πn⇒x/2=-π/3+2π/3+2πn=π/3+2πn⇒x=2π/3+4πn
x/2+π/3=-2π/3+2πn⇒x/2=-π/3-2π/3+2πn=-π+2πn⇒x=-2π+4πn