Question

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

Answer 1

Ответ:

$x=dfrac{pi}{4}+npi,; nin Z\\x=2npi,; nin Z\\x=dfrac{pi}{2}+2npi,; nin Z$

Пошаговое объяснение:

Область определения тангенса:

$x-pi/4ne pi/2+npi,; nin Z\xne3pi/4+npi,; nin Z$

Перейдем к решению:

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

Теперь нужно решить 2 уравнения:

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

И еще:

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

Answer 2

Сейчас допишу решение

Answer 3

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

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

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

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

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

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

Область допустимых значений (ОДЗ):

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

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

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

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

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

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

Решим уравнение $(2)$:

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

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

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

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

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

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

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

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

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