Решите уравнения, преобразовав их к однородным относительно синуса и косинуса:
[tex]6 \sin^{2} ( \frac{x}{2} - \frac{\pi}{6} ) + 0.5\sin( x - \frac{\pi}{3} ) = 2 + \cos^{2} ( \frac{\pi}{6} - \frac{x}{2} ) \\ \\ 8 { \sin( \frac{\pi}{8} - 2x) }^{2} - { \cos(2x - \frac{\pi}{8} ) }^{2} = 0.5 \sin(4x - \frac{\pi}{4} ) + 3[/tex]

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)=04sin²(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=0tg(x/2-π/6)=a4a²+a-3=0D=1+48=49a1=(-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∈za2=(-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∈z28sin²(2x-π/8)-cos²(2x-π/8)-sin(2x-π/8)*cos(2x-π/8)-3sin²(2x-π/8)-3cos²(2x-π/8)=05sin²(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=0tg(2x-π/8)=a5a²-a-4=0D=1+80=81a1=(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∈za2=(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