Answer:
No solution.
Step-by-step explanation:
The given functions are
[tex]y=\cos2x[/tex] and [tex]y=1-\sin^2x[/tex].
To find the point of intersections of the graphs of the two functions: we equate them and solve for [tex]x[/tex].
[tex]\cos2x=1-\sin^2x[/tex]
Recall the double angle identity; [tex]\cos2x=cos^2x-sin^2x[/tex]
Apply this identity to obtain;
[tex]cos^2x-sin^2x=1-\sin^2x[/tex]
[tex]\Rightarrow cos^2x=1[/tex]
[tex]\cos x=\pm1[/tex]
[tex]x=0\:or\:x=\pi[/tex]
if the interval is [tex]0\le x\le \pi[/tex], then the two graphs intersect at [tex]x=0\:or\:x=\pi[/tex]
But [tex]x=0\:and\:x=\pi[/tex] does not belong to the open interval [tex]0\:<\:x\:<\:\pi[/tex]
No point of intersection.
