Hei, har gjort en oppgave her, men jeg fikk ikke det samme svaret som fasiten, håper noen kan hjelpe meg.
Løs likningen for x e [0, 365] (fasit: x= 240)
([symbol:rot] (3)/2)*Sinx + (1/2)*cosX = -1
Trigonometrisk likning 3MX
Moderatorer: Aleks855, Gustav, Nebuchadnezzar, Janhaa, DennisChristensen, Emilga
[tex]\frac{\sqrt{3}}{2} \sin (x) + \frac{1}{2} \cos (x) = -1[/tex]
[tex]\frac{\sqrt{3}}{2} \sin (x) + \frac{1}{2} \cos (x) = \sqrt{ \left (\frac{\sqrt{3}}{2} \right )^2 + \left (\frac{1}{2} \right)^2} \cdot \sin (x + \phi)[/tex]
[tex]= \sqrt{\frac{3}{4} + \frac{1}{4}} \sin (x + \phi)[/tex]
[tex]= \sin (x + \phi)[/tex]
[tex]\tan \phi = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}[/tex]
[tex]\phi = 30^\circ[/tex] (siden [tex]\phi[/tex] skal ligge i 1. kvadrant i dette tilfellet)
[tex]\frac{\sqrt{3}}{2} \sin (x) + \frac{1}{2} \cos (x) = \sin (x + 30^\circ)[/tex]
[tex]\sin (x + 30^\circ) = -1[/tex]
[tex]x + 30^\circ = 270^\circ[/tex]
[tex]x = 240^\circ[/tex]
[tex]\frac{\sqrt{3}}{2} \sin (x) + \frac{1}{2} \cos (x) = \sqrt{ \left (\frac{\sqrt{3}}{2} \right )^2 + \left (\frac{1}{2} \right)^2} \cdot \sin (x + \phi)[/tex]
[tex]= \sqrt{\frac{3}{4} + \frac{1}{4}} \sin (x + \phi)[/tex]
[tex]= \sin (x + \phi)[/tex]
[tex]\tan \phi = \frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}} = \frac{1}{\sqrt{3}}[/tex]
[tex]\phi = 30^\circ[/tex] (siden [tex]\phi[/tex] skal ligge i 1. kvadrant i dette tilfellet)
[tex]\frac{\sqrt{3}}{2} \sin (x) + \frac{1}{2} \cos (x) = \sin (x + 30^\circ)[/tex]
[tex]\sin (x + 30^\circ) = -1[/tex]
[tex]x + 30^\circ = 270^\circ[/tex]
[tex]x = 240^\circ[/tex]