Question

Rozwiąż równanie trygonometryczne

Answer 1

Odpowiedź:

[tex]x\in\left\{0,\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4},2\pi\right\}[/tex]

Szczegółowe wyjaśnienie:

[tex]\sin x+\cos x=\frac{\cos^2x-\sin^2x}{1-2\sin x\cos x}\qquad x\in[0,2\pi][/tex]

Założenie:

[tex]1-2\sin x\cos x\neq 0\\-2\sin x\cos x\neq -1\ |:(-1)\\2\sin x\cos x\neq 1\\\sin(2x)\neq 1\\2x\neq \frac{\pi}{2}+2k\pi\ |:2\\x\neq \frac{\pi}{4}+k\pi\quad\text{ale }x\in[0,2\pi]\\x\neq \frac{\pi}{4}\ \land \ x\neq \frac{5\pi}{4}\\D=\left[0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right][/tex]

Rozwiązanie:

[tex]\sin x+\cos x=\frac{\cos^2x-\sin^2x}{1-2\sin x\cos x}\\\\\sin x+\cos x-\frac{\cos^2x-\sin^2x}{1-2\sin x\cos x}=0\\\\\sin x+\cos x-\frac{(\cos x-\sin x)(\cos x+\sin x)}{1-2\sin x\cos x}=0\\\\(\sin x+\cos x)\left(1-\frac{\cos x-\sin x}{1-2\sin x\cos x}\right)=0\\\\(\sin x+\cos x)\left(\frac{1-2\sin x\cos x}{1-2\sin x\cos x}-\frac{\cos x-\sin x}{1-2\sin x\cos x}\right)=0\\\\(\sin x+\cos x)\frac{1-2\sin x\cos x-\cos x+\sin x}{1-2\sin x\cos x}=0[/tex]

[tex](\sin x+\cos x)\frac{\sin^2x+\cos^2x-2\sin x\cos x+\sin x-\cos x}{1-2\sin x\cos x}=0\\\\(\sin x+\cos x)\frac{(\sin x-\cos x)^2+(\sin x-\cos x)}{1-2\sin x\cos x}=0\\\\(\sin x+\cos x)\frac{(\sin x-\cos x)(\sin x-\cos x+1)}{1-2\sin x\cos x}=0\ |*(1-2\sin x\cos x)\\\\(\sin x+\cos x)(\sin x-\cos x)(\sin x-\cos x+1)=0\\\\(\sin x+\cos x=0)\ \vee\ (\sin x-\cos x=0)\ \vee\ (\sin x-\cos x+1=0)[/tex]

Dla pierwszego nawiasu:

[tex]\sin x+\cos x=0\\\sin x=-\cos x\\\sin x=-\sin \left(\frac{\pi}{2}+x\right)\\\sin x=\sin \left(-\frac{\pi}{2}-x\right)\\x=-\frac{\pi}{2}-x+2k\pi\\2x=-\frac{\pi}{2}+2k\pi\ |:2\\x=-\frac{\pi}{4}+k\pi\quad\text{ale }x\in\left[0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right]\\x\in\left\{\frac{3\pi}{4},\frac{7\pi}{4}\right\}[/tex]

Dla drugiego nawiasu:

[tex]\sin x-\cos x=0\\\sin x=\cos x\\x=\frac{\pi}{4}+2k\pi\quad\text{ale }x\in\left[0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right]\\x\in\emptyset[/tex]

Dla trzeciego nawiasu:

[tex]\sin x-\cos x+1=0\\\sin x=\cos x-1\\\sin^2x+\cos^2x=1\\(\cos x-1)^2+\cos^2x=1\\\cos^2x-2\cos x+1+\cos^2x=1\\2\cos^2x-2\cos x=0\ |:2\\\cos^2x-\cos x=0\\\cos x(\cos x-1)=0\\\left \{ {{\cos x=0} \atop {\sin x=-1}} \right. \ \vee\ \left \{ {{\cos x=1} \atop {\sin x=0}} \right. \\x=\frac{3\pi}{2}+2k\pi\ \vee\ x=2k\pi\quad\text{ale }x\in\left[0,\frac{\pi}{4}\right)\cup\left(\frac{\pi}{4},\frac{5\pi}{4}\right)\cup\left(\frac{5\pi}{4},2\pi\right]\\x\in\left\{0,\frac{3\pi}{2},2\pi\right\}[/tex]

Ostatecznie:

[tex]x\in\left\{0,\frac{3\pi}{4},\frac{3\pi}{2},\frac{7\pi}{4},2\pi\right\}[/tex]