Question

Вторая задача. Решите 50 баллов. Если не так бан. Укажите решение

Answer 1

$\displaystyle sin^2x+sin^22x=sin^23x\\sin^2x+(2sinx*cosx)^2=(3sinx-4sin^3x)^2\\sin^2x+4sin^2x*cos^2x=9sin^2x-24sin^4x+16sin^6x\\sin^2x+4sin^2x(1-sin^2x)=9sin^2x-24sin^4x+16sin^6x\\sin^2x+4sin^2x-4sin^4x=9sin^2x-24sin^4x+16sin^6x\\0=16sin^6x-20sin^4x+4sin^2x\\4sin^2x(4sin^4x-5sin^2x+1)=0\\\left \{ {{4sin^2x=0} \atop {4sin^4x-5sin^2x+1=0}} \right.$

1)

$\displaystyle 4sin^2x=0\\sin^2x=0; sinx=0; x=\pi n; n\in Z$

2)

$\displaystyle 4sin^4x-5sin^2x+1=0\\sin^2x=t\\4t^2-5t+1=0\\D=25-19=9\\t_{1.2}=\frac{5 (+/-) 3}{8} \\t_1=1; t_2=1/4$

$\displaystyle sin^2=1; sinx= \pm 1; x=\frac{\pi}{2}+\pi n; n\in Z$

$\displaystyle sin^2x=\frac{1}{4}; sinx= \pm\frac{1}{2}\\x=\pm\frac{\pi }{6}+2\pi n; n\in Z\\x=\pm\frac{5\pi }{6}+2\pi n; n\in Z$