Question

sinx+cosx+sinx*cosx=1

Answer 1

Sinx + Cosx + Sinx * Cosx = 1

Обозничим Sinx + Cosx = m

Возведём обе части полученного равенства в квадрат :

(Sinx + Cosx)² = m²

Sin²x + 2SinxCosx + Cos²x = m²

Так как Sin²x + Cos²x = 1 , то :

1 + 2SinxCosx = m²

2SinxCosx = m² - 1

$SinxCosx=\frac{m^{2}-1 }{2}\\\\m+\frac{m^{2}-1 }{2}=1|*2\\\\2m+m^{2}-1=2 \\\\m^{2}+2m-3=0\\\\m_{1}=1\\\\m_{2}=-3$

1) Sinx + Cosx = - 3 - решений  нет, так как |Sinx| ≤ 1 и |Cosx| ≤ 1

$2)Sinx+Cosx=1|:\sqrt{2} \\\\\frac{1}{\sqrt{2}}Sinx+\frac{1}{\sqrt{2}}Cosx=\frac{1}{\sqrt{2}}\\\\Cos\frac{\pi }{4}Sinx+Sin\frac{\pi }{4}Cosx=\frac{1}{\sqrt{2}}\\\\Sin(x+\frac{\pi }{4})=\frac{1}{\sqrt{2}}$

$x+\frac{\pi }{4} =(-1)^{n}\frac{\pi }{4}+\pi n,n\in Z\\\\x=(-1)^{n}\frac{\pi }{4}-\frac{\pi }{4}+\pi n,n\in Z\\\\Otvet:\boxed{(-1)^{n}\frac{\pi }{4}-\frac{\pi }{4}+\pi n,n\in z}$

Answer 2

Ответ:   $x=(-1)^{n}\cdot \frac{\pi}{4}-\frac{\pi}{4}+\pi n\; ,\; n\in Z\; .$

$sinx+cosx+sinx\cdot cosx=1\\\\\\t=sinx+cosx\; \; \to \; \; t^2=(sinx+cosx)^2=sin^2x+cos^2x+2sinx\cdot cosx\; ,\\\\t^2=1+2sinx\cdot cosx\; \; \to \; \; sinx\cdot cosx=\frac{t^2-1}{2}\\\\t+\frac{t^2-1}{2}=1\; \; ,\; \; 2t+t^2-1=2\; \; ,\; \; t^2+2t-3=0\; ,\; \; \\\\t_1=1\; \; ,\; \; t_2=-3\; ,\\\\\\a)\; \; sinx+cosx\ne -3\; ,\; t.k.\; \; |sinx|\leq 1\; \; i\; \; |cosx|\leq 1\\\\b)\; \; sinx+cosx=1\; |:\sqrt2\\\\\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}\, cosx=\frac{1}{\sqrt2}\; \; \; \; \to \; \; \; cos\frac{\pi}{4}\, sinx+sin\frac{\pi}{4}\, cosx=\frac{\sqrt2}{2}\\\\sin(x+\frac{\pi}{4})=\frac{\sqrt2}{2}$

$x+\frac{\pi}{4}=(-1)^{n}\cdot \frac{\pi}{4}+\pi n\; ,\; n\in Z\\\\x=(-1)^{n}\cdot \frac{\pi}{4}-\frac{\pi}{4}+\pi n=\left \{ {{-\frac{\pi}{2}+\pi n\; ,\; esli\; n=2k-1\; ,\; k\in Z\; ,} \atop {\pi n\; ,\; esli\; n=2k\; ,\; k\in Z\; .\; \; \; \; \; }} \right.\; .$