Hvordan løse trigonometriske likninger

$$\begin{array}{llll}\hfill 4\cos 2x+4\sin 2x& =0|:4& \hfill & \\ \hfill \cos 2x+\sin 2x& =0|:\cos 2x\ne 0& \hfill & \\ \hfill \frac{\cos 2x}{\cos 2x}+\frac{\sin 2x}{\cos 2x}& =0& \hfill & \\ \hfill 1+\tan 2x& =0& \hfill & \\ \hfill \tan 2x& =-1& \hfill & \\ \hfill 2x& =\frac{3\pi }{4}+n\cdot \pi & \hfill & \\ \hfill x& =\frac{3\pi }{8}+n\cdot \frac{\pi }{2}& \hfill & \end{array}$$ I intervallet $[0,4\pi ⟩$ gir det løsningene Du må nå sjekke om $\cos 2x=0$ gir flere løsninger. Merk at $\cos 2x=0$ er det samme som $\sin 2x=1$ eller $\sin 2x=-1$: Fra dette får du følgende løsninger i intervallet $[0,4\pi ⟩$:

$$x\in \left\{\frac{\pi }{4},\frac{3\pi }{4},\frac{5\pi }{4},\frac{7\pi }{4},\frac{9\pi }{4},\frac{11\pi }{4},\frac{13\pi }{4},\frac{15\pi }{4}\right\}$$

Du må nå sette prøve på disse svarene i hovedlikningen (1):

$x=\frac{\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{\pi }{4}\right)+4\sin \left(2\cdot \frac{\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{3\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{3\pi }{4})\right)+4\sin \left(2\cdot \frac{3\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{5\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{5\pi }{4}\right)+4\sin \left(2\cdot \frac{5\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{7\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{7\pi }{4}\right)+4\sin \left(2\cdot \frac{7\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{9\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{9\pi }{4}\right)+4\sin \left(2\cdot \frac{9\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{11\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{11\pi }{4}\right)+4\sin \left(2\cdot \frac{11\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{13\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{13\pi }{4}\right)+4\sin \left(2\cdot \frac{13\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot 1& \hfill & \\ \hfill & =4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

$x=\frac{15\pi }{4}$:

$$\begin{array}{llll}\hfill \text{V.S}& =4\cos \left(2\cdot \frac{15\pi }{4}\right)+4\sin \left(2\cdot \frac{15\pi }{4}\right)& \hfill & \\ \hfill & =4\cdot 0+4\cdot (-1)& \hfill & \\ \hfill & =-4& \hfill & \\ \hfill & \ne 0& \hfill & \\ \hfill & =\text{H.S}& \hfill & \end{array}$$

Likningen (1) har ingen flere løsninger og du kan konkludere med at løsningsmengden i intervallet $[0,4\pi ⟩$ er:

$$x\in \left\{\frac{3\pi }{8},\frac{7\pi }{8},\frac{11\pi }{8},\frac{15\pi }{8},\frac{19\pi }{8},\frac{23\pi }{8},\frac{27\pi }{8},\frac{31\pi }{8}\right\}$$

NB! Dette er en svært tungvint måte å sjekke om $\cos 2x=0$ gir flere løsninger, men den fører alltid frem! En annen måte å sjekke $\cos 2x=0$ er å betrakte hovedlikningen og bruke det du kan om trigonometriske funksjoner til å analysere.

$$\begin{array}{llll}\hfill \cos (2x)+\cos x+1& =0& \hfill & \\ \hfill {\cos }^{2}x-{\sin }^{2}x+\cos x+1& =0& \hfill & \\ \hfill {\cos }^{2}x-\left(1-{\cos }^{2}x\right)+\cos x+1& =0& \hfill & \\ \hfill 2{\cos }^{2}x+\cos x& =0& \hfill & \\ \hfill \cos x(2\cos x+1)& =0& \hfill & \end{array}$$ Du bruker nullfaktorregelen og ser på de to likningene $\cos x=0$ og $2\cos x+1=0$. Likningen $\cos x=0$ har løsningene $$\begin{array}{llll}\hfill x_1& =\frac{\pi }{2}+n\cdot 2\pi ,& \hfill & \\ \hfill x_2& =-\frac{\pi }{2}+n\cdot 2\pi .& \hfill & \end{array}$$

Likingen $2\cos x+1=0$ skriver du om som grunnlikningen $\cos x=-\frac{1}{2}$. Den har løsningene

$$\begin{array}{llll}\hfill x_3& =\frac{2\pi }{3}+n\cdot 2\pi ,& \hfill & \\ \hfill x_4& =-\frac{2\pi }{3}+n\cdot 2\pi .& \hfill & \end{array}$$

Løsningsmengden for intervallet $[0,2\pi ⟩$ blir dermed

$$x\in \left\{\frac{\pi }{2},\frac{2\pi }{3},\frac{4\pi }{3},\frac{3\pi }{2}\right\}.$$

$$\begin{array}{llll}\hfill 2\left(1-{\sin }^{2}x\right)-\sin x& =1& \hfill & \\ \hfill 2-2{\sin }^{2}x-\sin x& =1& \hfill & \\ \hfill -2{\sin }^{2}x-\sin x+1& =0& \hfill & \end{array}$$ Du setter $u=\sin x$ inn i likningen og får:

$$-2u^2-u+1=0.$$

Løs andregradslikningen:

$$u=\frac{1\pm \sqrt{{(-1)}^2-4\cdot (-2)\cdot 1}}{-4}=\frac{1\pm \sqrt{9}}{-4}$$

Altså er løsningene

$$u_1=-1u_2=\frac{1}{2}$$

Sett nå $\sin x$ inn for $u$:

Løsningsmengden på intervallet $[0,2\pi ⟩$ blir dermed:

$$x\in \left\{\frac{\pi }{6},\frac{5\pi }{6},\frac{3\pi }{2}\right\}$$

Legg merke til at $-\frac{\pi }{2}$ ikke er med i løsningsmengden. Det er fordi den ligger utenfor intervallet du undersøker.