Решить задания(предпочтительнее первый вариант)

1 1)cosx=1/√2 x=+-π/4+2πk,k∈z 2)tg2x=-√3/3 2x=-π/6+2πk x=-π/12+πk/2,k∈z 2 x/3=7π/6+2πk,k∈z U x=11π/6+2πn,n∈z 0≤7π/6+2πk≤3π 0≤7+12k≤18 -7≤12k≤11 -7/12≤k≤11/12 k=0⇒x=7π/6 0≤11π/6+2πn≤3π 0≤11+12n≤18 -11≤12n≤7 -11/12≤n≤7/12 n=0⇒x=11π/6 3 1)cosx(3-cosx)=0 cosx=0⇒x=π/2+πk,k∈z cosx=3>1 нет решения 2)sinx=a 6a²-a-1=0 D=1+24=25 a1=(1-5)/12=-1/3⇒sinx=-1/3⇒x=(-1)^(n+1)*arcsin1/3+πn,n∈z a2=(1+5)/12=1/2⇒sinx=1/2⇒x=(-1)^k*π/6+πk,k∈z 3)8sin(x/2)cos(x/2)+5cos²(x/2)-5sin²(x/2)-4sin²(x/2)-4cos²(x/2)=0/cos²(x/2) 9tg²(x/2)-8tg(x/2)-1=0 tg(x/2)=a 9a²-8a-1=0 D=64+36=100 a1=(8-10)/18=-1/9⇒tg(x/2)=-1/9⇒x/2=-arctg1/9+πn⇒ x=-2arctg1/9+2πn,n∈z a2=(8+10)/18=1⇒tg(x/2)=1⇒x/2=π/4+πk⇒x=π/2+2πk,k∈z 4)(1-cos2x)²/4+(1+cos2x)²/4=cos²2x+1/4 1-2cos2x+cos²2x+1+2cos2x+cos²2x=4cos²2x+1 2+2cos²2x-4cos²2x-1=0 2cos²2x-1=0 cos4x=0 4x=π/2+πk,k∈z x=π/8+πk/4,k∈z