Question

Решите тригонометрические уравнения. Хотя бы одно!

cos(7X)-cos(X)-sin(4X)=0
sin^2(X)+6cos^2(X)+7sin(X)cos(X)=0
4sin^2(X)+5sin(X)cos(X)-cos^2(X)=2
sin(2X)+корень из 2* sin(x-п/4)=1

Answer 1

$cos (7x) -cos x - sin (4x)=0\\-2cdot sindfrac{7x+x}2cdot sin dfrac{7x-x}2-sin(4x)=0\\-2cdot sin(4x)cdot sin (3x)-sin(4x)=0~~~~|cdot (-1)\2cdot sin(4x)cdot sin (3x)+sin(4x)=0\sin(4x)Big(2sin (3x)+1Big)=0$

$1) ~~sin(4x)=0;\~~~~~4x=pi n,~~x_1=dfrac{pi n}4,~~n in Z\\2) ~~2sin(3x)+1=0;~~sin (3x)=-dfrac12\\~~~~left[begin{array}{c}3x=-dfrac{pi }6+2pi k;~~x_2=-dfrac{pi}{18}+dfrac{2pi k}3,~~k in Z\\3x=-dfrac{5pi }6+2pi m;~~x_3=-dfrac{5pi}{18}+dfrac{2pi m}3,~~m in Zend{array}$

Ответ:  $dfrac{pi n}4;~~~~-dfrac{pi}{18}+dfrac{2pi k}3;~~~~-dfrac{5pi}{18}+dfrac{2pi m}3,~~n,k,m in Z$

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$sin^2x+6cos^2x+7sin x cos x=0~~~~|:cos x neq 0~\\ dfrac{sin^2x}{cos^2x}+6+dfrac{7sin xcos x}{cos^2 x}=0\\ tg^2x+7~tgx + 6 = 0\(tgx+6)(tgx+1)=0\\1)~tg x = -6; ~~x_1=-arctg~6+pi n,~~nin Z\\2)~tgx=-1;~~x_2=-dfrac{pi}4+pi k,~~k in Z$

Ответ: $-arctg~6+pi n;~~~-dfrac{pi}4+pi k,~~n,k in Z$

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$4sin^2x+5sin xcos x-cos^2x=2\4sin^2x+5sin xcos x-cos^2x-2(sin^2x+cos^2x)=0\4sin^2x+5sin xcos x-cos^2x-2sin^2x-2cos^2x=0\2sin^2x+5sin xcos x-3cos^2x=0~~~~|:cos^2xneq 0\\dfrac{2sin^2x}{cos^2x}+dfrac{5sin xcos x}{cos^2x}-dfrac{3cos^2x}{cos^2x}=0\\2tg^2x+5tgx-3=0\D=25-4cdot 2cdot (-3)=49=7^2\\1)~tgx=dfrac{-5+7}4=dfrac12;~~x_1=arctgdfrac12+pi n,~n in Z\\2)~tgx=dfrac{-5-7}4=-3;~~x_2=-arctg3+pi k,~k in Z$

Ответ: $arctgdfrac12+pi n;~~~-arctg3+pi k,~n,k in Z$

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$sin (2x) + sqrt 2sin Big(x-dfrac{pi}4Big)=1\\ 2sin xcos x + sqrt 2Big(sin xcos dfrac{pi}4-sindfrac{pi}4cos xBig)=1\\ 2sin xcos x + sqrt 2Big(dfrac{sqrt2}2sin x-dfrac{sqrt2}2cos xBig)=1\\ 2sin xcos x + sin x-cos x-1=0\2sin xcos x + sin x-cos x-sin^2-cos^2x=0~~~|cdot (-1)\ cos^2x-2sin xcos x + sin^2x+cos x-sin x=0\(cos x- sin x)^2+(cos x-sin x)=0\(cos x- sin x)(cos x-sin x+1)=0$

$1)~cos x-sin x=0;~~cos x=sin x\\~~~x_1=dfrac{pi}4+pi n,~n in Z\\2)~cos x-sin x+1=0;~~(1+cos x)-sin x=0\\~~~~2cos^2dfrac x2-2sin dfrac x2cos dfrac x2=0\\ ~~~~2cos dfrac x2Big(cos dfrac x2-sin dfrac x2Big)=0\\~~~~a) cos dfrac x2=0;~~dfrac x2=dfrac {pi}2+pi k;~~x_2=pi + 2pi k,~k in Z\\~~~~b)cos dfrac x2-sin dfrac x2=0;~~cos dfrac x2=sin dfrac x2\\~~~~dfrac x2=dfrac {pi}4+ pi m; ~~x_3=dfrac{pi}2+2pi m, ~m in Z$

Ответ: $dfrac{pi}4+pi n;~~pi + 2pi k;~~dfrac{pi}2+2pi m, ~~n,k,m in Z$