Ответы на вопрос:
16sin²(x/2-π/6)+sin(x/2-π/6)*cos(x/2-π/6)-2sin²(x/2-π/6)-2cos²(x/2-π/6)-cos²(x/2-π/6)=0 4sin²(x/2-π/6)+sin(x/2-π/6)*cos(x/2-π/6)-3cos²(x/2-π/6)=0 /cos²(x/2-π/6) 4tg²(x/2-π/6)+tg(x/2-π/6)-3=0 tg(x/2-π/6)=a 4a²+a-3=0 d=1+48=49 a1=(-1-7)/8=-1⇒tg(x/2-π/6)=-1⇒x/2-π/6=-π/4+πk⇒x/2=-π/12+πk⇒ x=-π/6+2πk,k∈z a2=(-1+7)/8=3/4⇒tg(x/2-π/6)=3/4⇒x/2-π/6=arctg3/4+πk⇒ x/2=π/6+arctg3/4+πk⇒x=π/3+2arctg3/4+2πk,k∈z 2 8sin²(2x-π/8)-cos²(2x-π/8)-sin(2x-π/8)*cos(2x-π/8)-3sin²(2x-π/8)-3cos²(2x-π/8)=0 5sin²(2x-π/8)-sin(2x-π/8)*cos(2x-π/8)-4cos²(2x-π/8)=0 /cos²(2x-π/8) 5tg²(2x-π/8)-tg(2x-π/8)-4=0 tg(2x-π/8)=a 5a²-a-4=0 d=1+80=81 a1=(1-9)/10=-0,8⇒tg(2x-π/8)=-0,8⇒2x-π/8=-arctg0,8+πk⇒ 2x=π/8-arctg0,8+πk⇒x=π/16-0,5arctg0,8+πk/2,k∈z a2=(1+9)/10=1⇒tg(2x-π/8)=1⇒2x-π/8=π/4+πk⇒2x=3π/8+πk⇒ x=3π/16+πk/2,k∈z
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