Question

Помогите решить, а то не получается

Answer 1

$(Sinx - Cosx) ^{2}= \frac{3- \sqrt{3} }{2}Cos ^{2}2x\\\\Sin ^{2}x -2SinxCosx+Cos ^{2}x= \frac{3- \sqrt{3} }{2}Cos ^{2} 2x\\\\1-Sin2x = \frac{3- \sqrt{3} }{2}(1-Sin ^{2} 2x)\\\\1-Sin2x= \frac{3- \sqrt{3} }{2}- \frac{3- \sqrt{3} }{2}Sin ^{2} 2x\\\\2-2Sin2x=3- \sqrt{3}-(3- \sqrt{3})Sin ^{2}2x \\\\(3- \sqrt{3})Sin ^{2}2x-2Sin2x+2-3+ \sqrt{3} =0\\\\(3- \sqrt{3})Sin ^{2}2x-2Sin2x+( \sqrt{3}-1)=0\\\\D=4-4(3- \sqrt{3})( \sqrt{3} -1)=4-4(3 \sqrt{3}-3-3+ \sqrt{3}) }=$$4-4(4 \sqrt{3}-6)=4-16 \sqrt{3}+24=28-16 \sqrt{3}=4(7-4 \sqrt{3})=$$[2(2- \sqrt{3} )] ^{2}\\\\Sin2x= \frac{2+2(2- \sqrt{3}) }{2(3- \sqrt{3}) }= \frac{6-2 \sqrt{3} }{6-2 \sqrt{3} }=1\\\\2x= \frac{ \pi }{2} +2 \pi n\\\\ x_{1} = \frac{ \pi }{4} + \pi n\\\\Sin2x= \frac{2-2(2- \sqrt{3}) }{2(3- \sqrt{3}) } = \frac{2-4+2 \sqrt{3} }{2(3- \sqrt{3}) } = \frac{2 \sqrt{3} -2}{2(3- \sqrt{3}) } = \frac{2( \sqrt{3}-1) }{2 \sqrt{3}( \sqrt{3}-1) } = \frac{1}{ \sqrt{3} }\\\\2x=(-1) ^{n}arcSin \frac{1}{ \sqrt{3} } + \pi n\\\\ x_{2} =(-1) ^{n} \frac{1}{2}arcSin \frac{1}{ \sqrt{3} } + \frac{ \pi n}{2}$
Сумма корней $arcSin \frac{1}{ \sqrt{3} } + \frac{ \pi }{4}$