Question

Решите,,уравнение: 5sin2x-11(sinx+cosx)+7=0

Answer 1

$5sin2x-11(sinx+cosx)+7=0\\\\t=sinx+cosx\; ,t^2=sin^2x+cos^2x+2sinx\cdot cosx=1+sin2x\; \Rightarrow \\\\sin2x=t^2-1\\\\5(t^2-1)-11t+7=0\\\\5t^2-11t+2=0\\\\D=121-40=81\\\\t_1=\frac{11-9}{10}=\frac{2}{10}=\frac{1}{5}\; ,\; t_2=\frac{20}{10}=2\\\\a)\; \; sinx+cosx=\frac{1}{5}|:\sqrt2$

$\frac{1}{\sqrt2}sinx+\frac{1}{\sqrt2}cosx=\frac{1}{5\sqrt2}\\\\cos\frac{\pi}{4}sinx+sin\frac{\pi}{4}cosx=\frac{1}{5\sqrt2}\\\\sin(x+\frac{\pi}{4})=\frac{1}{5\sqrt2}\\\\x+\frac{\pi}{4}+(-1)^{n}\cdot arcsin\frac{1}{5\sqrt2}+\pi n\; ,\; n\in Z\\\\x=(-1)^{n}\cdot arcsin \frac{1}{5\sqrt2}-\frac{\pi}{4}+\pi n\; ,\; n\in Z\\\\b)\; \; sinx+cosx=2\\\\sin(x+\frac{\pi}{4})=\frac{2}{\sqrt2}\ \textgreater \ 1\; \; \Rightarrow \; \; ney\; reshenij,\; t.k.\; |sin \alpha | \leq 1$

ответ:  $x=(-1)^{n}arcsin\frac{1}{5\sqrt2}-\frac{\pi}{4}+\pi n,\; n\inZ$

Можно было преобразовать sinx+cosx по другому:

$sinx+cosx=sinx+sin(\frac{\pi}{2}-x)=2sin\frac{x+(\frac{\pi }{2}-x)}{2}\cdot cos\frac{x-(\frac{\pi}{2}-x)}{2}=\\\\=2sin\frac{\pi}{4}cos(x-\frac{\pi}{4})=2\cdot \frac{1}{\sqrt2}cos(x-\frac{\pi}{4})=\sqrt2cos(x-\frac{\pi}{4})$

Тогда 

 $\sqrt2cos(x-\frac{\pi}{4})=\frac{1}{5}\\\\cos(x-\frac{\pi}{4})=\frac{1}{5\sqrt2}$

$x-\frac{\pi}{4}=\pm arccos\frac{1}{5\sqrt2}+2\pi n\; ,\; n\in Z\\\\x=\frac{\pi}{4}\pm arccos\frac{1}{5\sqrt2}+2\pi n\; ,\; n\in Z$

ответ будет иметь другой вид, но это те же точки.