\(x=\pi+k\cdot2\pi\)

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We are actually looking for the intersections of the graph of cosine function with the line \(y=-1\). We make use of the unit circle because then the problem can be translated into the finding of points of intersection of the unit circle with the #x#-axis and into searching the angles of rotation that move the point #A=(1,0)# on the unit circle to these points of intersection.

The point #(-1,0)# is the intersection point from the unit circle with the #x#-axis dat verkregen wordt by truning the point #A=(1,0)# turning over an angle #\pi#. But there are many more rotation angles about #A# move there. Be an integral multiple of at each rotation angle \(2\pi\) add or subtract. This gives #x=\pi+k\cdot2\pi# surely integer \(k\) . so:

The solution of the equation \(\cos x=-1\) is of the form \(x=\pi+k\cdot2\pi\) with \(k\) an arbitrary integer.

Warning
The temptation is great on both sides of the equation \(\cos(x)=-1\) function #\arccos# apply. You get \(\arccos\bigl(\cos(x)\bigr)=\arccos(-1)\) and you might think you can use that #\cos# and #\arccos# inverses each other and that therefore common #x=\arccos(-1)=\pi# .
But then you've just found one solution! This is because #\arccos\bigl(\cos(x)\bigr)=x# only applies to the segment #[0,\pi]# and have you found a solution in this segment. You have to figure out what multiple of #\pi# may be hereby added or subtracted.