Respuesta :
first of all, we must use the equivalent formula of cos2x, it is cos2x=2cos^2x-1,
the equation has become 2cos^2x-1 +2cosx +1 = 0, it means 2cos^2x+2cosx = 0, or
cosx(cosx+1)=0, we get cosx=0, implying x=Pi/2, or (cosx+1)=0, cosx= -1
implying x = Pi
for x=0, cos2*0+2cos0+1=1+2+1=4 impossible
for x= Pi, cos2*Pi+2cosPi+1=1-2+1=0 it is exact , so the solution is x=Pi
the equation has become 2cos^2x-1 +2cosx +1 = 0, it means 2cos^2x+2cosx = 0, or
cosx(cosx+1)=0, we get cosx=0, implying x=Pi/2, or (cosx+1)=0, cosx= -1
implying x = Pi
for x=0, cos2*0+2cos0+1=1+2+1=4 impossible
for x= Pi, cos2*Pi+2cosPi+1=1-2+1=0 it is exact , so the solution is x=Pi
Answer:
Here, the given expression is,
[tex]cos(2x)+2cosx+1=0[/tex]
[tex]\implies (2cos^2x-1)+2cosx+1=0[/tex] ( Because cos(2x) = 2cos²x-1 )
[tex]\implies 2cos^2x+2cosx=0[/tex]
[tex]\implies 2cosx(cosx+1)=0[/tex]
Thus, 2 cos x = 0 or cos x + 1 = 0
If ,
[tex]2cos x = 0[/tex]
⇒ [tex]cos x = 0[/tex]
⇒ [tex]x = \frac{pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}.......[/tex]
⇒ [tex]x = \frac{\pi}{2}+n\pi[/tex]
Where, n = 0, 1, 2, ......
Now, if,
[tex]cos x = -1[/tex]
⇒ [tex]x=\pi, 3\pi, 5\pi,.....[/tex]
⇒ [tex]x = \pi+n\pi[/tex]
Where, n = 0, 1, 2, ......