Question

Sinx+sin2x=cosx помогите пожалуйста, 10 класс

Answer 1

$sinx+sin2x=cosx\\\\(sinx-cosx)+sin2x=0\\\\\star \; \; t=sinx-cosx\; \; \Rightarrow \; \; t^2=(sinx-cosx)^2\\\\t^2=(sin^2x+cos^2x)-2sinx\cdot cosx=1-sin2x\; \; \Rightarrow \; \; sin2x=1-t^2\; \star \\\\t+(1-t^2)=0\; \; \Rightarrow \; \; t^2-t-1=0\; ,\; \; D=1+4=5\; ,\\\\t_{1,2}=\frac{1\pm \sqrt5}{2}\; ,\\\\a)\; \; sinx-cosx=\frac{1-\sqrt5}{2}\; ,\; \\\\\sqrt2\cdot (\frac{1}{\sqrt2}sinx-\frac{1}{\sqrt2}cosx)=\frac{1-\sqrt5}{2}\\\\\sqrt2\cdot (cos\frac{\pi}{4}sinx-sin\frac{\pi}{4}cosx)=\frac{1-\sqrt5}{2}\\\\\sqrt2\cdot sin(x-\frac{\pi}{4})=\frac{1-\sqrt5}{2}$

$sin(x-\frac{\pi }{4})=\frac{1-\sqrt5}{2\sqrt2} \approx -0,44\\\\x-\frac{\pi}{4}=(-1)^{n}\, arcsin\frac{1-\sqrt5}{2\sqrt2}+\pi n,\; n\in Z\\\\x=\frac{\pi }{4}+(-1)^{n}\, arcsin\frac{1-\sqrt5}{2\sqrt2}+\pi n=\frac{\pi}{4}+(-1)^{n+1}\, arcsin\frac{\sqrt5-1}{2\sqrt2}+\pi n,\; n\in Z\\\\b)\; \; sinx-cosx=\frac{1+\sqrt5}{2\sqrt2}\\\\sin(x-\frac{\pi}{4})=\frac{1+\sqrt5}{2\sqrt2}\approx 1,15>1\; \; \Rightarrow \; \; x\in \varnothing$

$Otvet:\; \; x=\frac{\pi }{4}+(-1)^{n+1}\, arcsin\frac{\sqrt5-1}{2\sqrt2}+\pi n,\; n\in Z\; .$