Question

помогите!! если лень решать, натолкните хотя бы на мысль ​

Answer 1

Ответ:

$sin^2(45^\circ +x)+cos^2(45^\circ -x)+cos^2(90^\circ +2x)=1\\\\(sin45^\circ \cdot cosx+cos45^\circ \cdot sinx)^2+(cos45^\circ \cdot cosx+sin45^\circ \cdot sinx)^2+(-sin2x)^2=1\\\\\Big(\dfrac{1}{\sqrt2}\cdot cosx+\dfrac{1}{\sqrt2}\cdot sinx\Big)^2+\Big(\dfrac{1}{\sqrt2}\cdot cosx+\dfrac{1}{\sqrt2}\cdot sinx\Big)^2+sin^22x=1\\\\2\cdot \Big(\dfrac{1}{\sqrt2}\cdot cosx+\dfrac{1}{\sqrt2}\cdot sinx\Big)^2+sin^22x=1\\\\2\cdot \dfrac{1}{2}\cdot \Big(cosx+sinx\Big)^2+sin^22x=1$

$2\cdot \dfrac{1}{2}\cdot \Big(\underbrace{cos^2x+sin^2x}_{1}+\underbrace{2sinx\cdot cosx}_{sin2x}\Big)+sin^22x=1$

$1+sin2x+sin^22x=1\\\\sin^22x+sin2x=0\\\\sin2x\, (sin2x+1)=0\\\\a)\ \ sin2x=0\ \ ,\ \ 2x=\pi n\ ,\ \ x=\dfrac{\pi n}{2}\ ,\ n\in Z\\\\b)\ \ sin2x=-1\ \ ,\ \ 2x=-\dfrac{\pi}{2}+2\pi n\ \ ,\ \ \ x=-\dfrac{\pi}{4}+\pi n\ ,\ n\in Z\\\\c)\ \ x\in [\, 0^\circ \ ;\, 180^\circ \ ]:\ \ x_1=0^\circ \ ,\ x_2=90^\circ \ ,\ x_3=135^\circ \ ,\ x_4=180^\circ \\\\x_1+x_2+x_3+x_4=405^\circ$