Даю 35 баллов. Помогите решить то что сможете.

1a)sin765=sin(720+45)=sin45=√2/2b)cos19π/6=cos(3π+π/6)=-cosπ/6=-√3/22cosa=-√(1-sin²a)=-√(1-0,09)=-√0,9131)cosacosb+sinasinb-cosacosb+sinasinb=2sinasinb2)(-sina-cosa)/(1-2cos²a)=(sina+cosa)/cos2a==(sina+cosa)/(cosa-sina)(cosa+sina)=1/(cosa-sina)4sinxcos3x+cosxsin3x=1sin(x+3x)=1sin4x=14x=π/2+2πn,n∈zx=π/8+πn/2,n∈z5(sina/cosa+cosa/sina)*2sin²2a=(sin²a+cos²a)/sinacosa*2sin²2a==2/sin2a*2sin²2a=4sin2a6a)sinx=1/√2x=(-1)^n*π/4+πn,n∈zb)tgx/2=√3x/2=π/3+πn,n∈zx=2π/3+2πn,n∈z7cosx/2=1/21)x/2=-π/3+2πn,n∈z⇒x=-2π/3+4πn0≤-2π/3+4πn≤4π0≤-2+12n≤122≤12n≤141/6≤n≤7/6n=1⇒x=-2π/3+4π=10π/32)x/2=π/3+2πk,k∈z⇒x=2π/3+4πk,k∈z0≤2π/3+4πk≤4π0≤2+12k≤12-2≤12k≤10-1/6≤k≤5/6 k=0⇒x=2π/381)sinx(sinx-1)=0sinx=0⇒x=πn,n∈zsinx=1⇒x=π/2+2πk,k∈z2)cosx=a10a²+3a-1=0D=9+40=49a1=(-3-7)/20=-1/2⇒cosx=-1/2⇒x=+-2π/3+2πn,n∈za2=(-3+7)/20=1/5⇒cosx=1/5⇒x=+-arccos1/5+2πk,k∈z3)10sinx/2cosx/2+cos²x/2-sin²x/2-5sin²x/2-5cos²x/2=0/cos²x/26tg²x/2-10tgx/2+4=0tgx/2=a6a²-10a+4=0D=100-96=4a1=(10-2)/12=2/3⇒tgx/2=2/3⇒x/2=arctg2/3+πn⇒x=2arctg2/3+πn,n∈za2=(10+2)/12=1⇒tgx/2=1⇒x/2=π/4+πk,k∈z⇒x=π/2+2πk,k∈z4)(1-cos2x)²/4+(1+cos2x)²/4=sin²2x1-2cos2x+cos²2x+1+2cos2x+cos²2x=16sin ²2x2+2cos²2x=16sin²2x1+cos²2x-8sin²2x=08(1-cos4x)/2-(1+cos4x)/2=18-8cos4x-1-cos4x=29cos4x=5cos4x=5/94x=+-arccos5/9+2πn,n∈zx=+-1/4*arccos5/9+πn/2,n∈z