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

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

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

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

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

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

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

ответ: $arctg\dfrac12+\pi n;~~~-arctg3+\pi k,~n,k \in Z$

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

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

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