Question

Помогите пожалуйста!

Answer 1

$frac{1+sqrt{3} }{2}*Sin2x=(sqrt{3}-1)Cos^{2}x+1\\frac{1+sqrt{3}}{2}*2SinxCosx=(sqrt{3}-1)Cos^{2}x+Sin^{2}x+Cos^{2}x\\(1+sqrt{3})SinxCosx=(sqrt{3}-1+1)Cos^{2}x+Sin^{2}x\\Sin^{2}x-(1+sqrt{3})SinxCosx+sqrt{3}Cos^{2}x=0|:Cos^{2}x,Cosxneq0\\tg^{2}x-(1+sqrt{3})tgx+sqrt{3}=0\\D=(1+sqrt{3})^{2}-4*sqrt{3}=1+2sqrt{3}+3-4sqrt{3}=4-2sqrt{3}=(1-sqrt{3})^{2}  \\tgx_{1}=frac{1+sqrt{3}+1-sqrt{3}}{2}=1$

$x_{1} =arctg1+pi n,nin Z\\x_{1}=frac{pi }{4} +pi n,nin Z\\tgx_{2}=frac{1+sqrt{3}-1+sqrt{3}}{2}=frac{2sqrt{3} }{2}=sqrt{3}\\x_{2}=arctgsqrt{3} +pi n,nin Z\\x_{2}=frac{pi }{3}+pi n,nin Z$

Answer 2

спасибо!

Answer 3

Здравствуйте, помогите, пожалуйста