РЕШИТЕ ПЖ, очень нужно 1)sin5χ=cos4χ 2) 2tgχ-2ctgχ=3 3)25sin²χ+100cosχ = 89

1)sin5χ=cos4χ sin5x-sin(π/2-4x)=0 2sin(9x/2-π/4)cos(x/2+π/4)=0 sin(9x/2-π/4)=0⇒9x/2-π/4=πn⇒9x/2=π/4+πn⇒x=π/18+2πn/9,n∈z cos(x/2+π/4)=0⇒x/2+π/4=π/2+πk⇒x/2=π/4+πk⇒x=π/2+2πk,k∈z 2) 2tgχ-2ctgχ=3 2tgx-2/tgx=3 2tg²x-3tgx-2=0,tgx≠0 tgx=a 2a²-3a-2=0 D=9+16=25 a1=(3-5)/4=-1/2⇒tgx=-1/2⇒x=-arctg1/2+πn,n∈z a2=(3+5)/4=2⇒tgx=2⇒x=arctg2+πk,k∈z 3)25sin²χ+100cosχ = 89 25-25cos²x+100cosx-89=0 cosx=a 25a²-100a+64=0 D=10000-6400=3600 a1=(100-60)/50=0,8⇒cosx=0,8⇒x=+-arccos0,8+2πn,n∈z a2=(100+60)/50=3,2⇒cosx=3,2>1 нет решения