Question

Помогите решить уравнение: tg(x-pi/4)=sinx-cosx

Answer 1

Ответ:

$x=\dfrac{\pi}{4}+n\pi,\; n\in Z\\\\x=2n\pi,\; n\in Z\\\\x=\dfrac{\pi}{2}+2n\pi,\; n\in Z$

Пошаговое объяснение:

Область определения тангенса:

$x-\pi/4\ne \pi/2+n\pi,\; n\in Z\\x\ne3\pi/4+n\pi,\; n\in Z$

Перейдем к решению:

$tg(x-\pi/4)=sinx-cosx\\\\sinx-cosx=\sqrt{2}(sinxcos\pi/4-sin\pi/4cosx)=\sqrt{2}sin(x-\pi/4)\\tg(x-\pi/4)=sin(x-\pi/4)/cos(x-\pi/4)\\\\=>\sin(x-\pi/4)/\cos(x-\pi/4)=\sqrt{2}\sin(x-\pi/4)\\\sin(x-\pi/4)=\sqrt{2}\sin(x-\pi/4)\times \cos(x-\pi/4)\\\sqrt{2}sin(x-\pi/4)\times \cos(x-\pi/4)-\sin(x-\pi/4)=0\\\sin(x-\pi/4)\times( \sqrt{2}\cos(x-\pi/4)-1)=0$

Теперь нужно решить 2 уравнения:

$\sin(x-\pi/4)=0\\x=\dfrac{\pi}{4}+n\pi,\; n\in Z$

И еще:

$\sqrt{2}\cos(x-\pi/4)-1=0\\\cos(x-\pi/4)=1/\sqrt{2}\\x=2n\pi,\; n\in Z\\x=\dfrac{\pi}{2}+2n\pi,\; n\in Z$

Answer 2

$\text{tg} \, \left(x - \dfrac{\pi}{4} \right) = \sin x - \cos x$

$\dfrac{\sin \left(x - \dfrac{\pi}{4} \right)}{\cos \left(x - \dfrac{\pi}{4} \right)} = \sin x - \cos x$

$\dfrac{\sin x \cos \dfrac{\pi}{4} - \cos x \sin \dfrac{\pi}{4} }{\cos x \cos \dfrac{\pi}{4}+ \sin x\sin \dfrac{\pi}{4} } = \sin x - \cos x$

$\dfrac{\dfrac{\sqrt{2}}{2}\sin x - \dfrac{\sqrt{2}}{2}\cos x }{\dfrac{\sqrt{2}}{2}\cos x + \dfrac{\sqrt{2}}{2}\sin x } = \sin x - \cos x$

$\dfrac{\dfrac{\sqrt{2}}{2}(\sin x - \cos x) }{\dfrac{\sqrt{2}}{2}(\cos x + \sin x)} = \sin x - \cos x$

$\dfrac{\sin x - \cos x}{\cos x + \sin x } = \sin x - \cos x$

Область допустимых значений (ОДЗ):

$\cos x + \sin x \neq 0\\\cos x \neq -\sin x \ \ \ | : \sin x \\\text{ctg} \, x \neq -1\\x \neq -\dfrac{3\pi}{4} + 2\pi n, \ n \in Z$

$\sin x - \cos x = (\sin x - \cos x)(\cos x + \sin x)$

$\sin x - \cos x - (\sin x - \cos x)(\cos x + \sin x) = 0$

$(\sin x - \cos x)(1 - \cos x - \sin x) = 0$

$\left[\begin{array}{ccc}\sin x - \cos x = 0 \ \ \ \ \ \ \ \ (1) \\1 - \cos x - \sin x = 0 \ \ \ (2)\\\end{array}\right$

Решим уравнение $(1)$:

$\sin x = \cos x \ \ \ | : \cos x \neq 0\\\text{tg} \, x = 1\\x = \dfrac{\pi}{4} + \pi n, \ n \in Z$

Так как $x \neq -\dfrac{3\pi}{4} + 2\pi n, \ n \in Z$, то $x = \dfrac{\pi}{4} + 2\pi n, \ n \in Z$

Решим уравнение $(2)$:

$\cos x + \sin x = 1 \ \ \ |:\sqrt{2}$

$\dfrac{1}{\sqrt{2}}\cos x + \dfrac{1}{\sqrt{2}} \sin x = \dfrac{1}{\sqrt{2}}$

$\dfrac{\sqrt{2}}{2}\cos x + \dfrac{\sqrt{2}}{2} \sin x = \dfrac{\sqrt{2}}{2}$

$\sin \dfrac{\pi}{4} \cos x + \cos \dfrac{\pi}{4}\sin x = \dfrac{\sqrt{2}}{2}$

$\sin \left(\dfrac{\pi}{4} + x \right)=\dfrac{\sqrt{2}}{2}$

$\left[\begin{array}{ccc}\sin \left(\dfrac{\pi}{4} + x \right)=\dfrac{\sqrt{2}}{2} \ \ \\\sin \left(\dfrac{3\pi}{4} - x \right)=\dfrac{\sqrt{2}}{2}\\\end{array}\right$

$\left[\begin{array}{ccc}\dfrac{\pi}{4} + x = \dfrac{\pi}{4} + 2\pi k, \ k \in Z \\ \\\dfrac{3\pi}{4} - x = \dfrac{\pi}{4} + 2\pi l, \ l \in Z \\\end{array}\right$

$\left[\begin{array}{ccc}x = 2\pi k, \ k \in Z \ \ \ \ \ \\x = \dfrac{\pi}{2} + 2\pi l, \ l \in Z \\\end{array}\right$

Ответ: $x = \dfrac{\pi}{4} + 2\pi n; \ x = 2\pi k; \ x = \dfrac{\pi}{2}+2\pi l; \ n,k,l \in Z$