Question

At what x-values do the graphs of the functions y=cos 2x and y=1-sin^2x intersect over the interval 0 < x < pi

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Answer 1

Answer: one of the answers pi/2 for the first half i dont know the other

Answer 2

No solution.

Explanation:

The given functions are

`y=\cos2x` and
`y=1-\sin^2x`.

To find the point of intersections of the graphs of the two functions: we equate them and solve for
`x`.

`\cos2x=1-\sin^2x`

Recall the double angle identity;
`\cos2x=cos^2x-sin^2x`

Apply this identity to obtain;

`cos^2x-sin^2x=1-\sin^2x`

`\Rightarrow cos^2x=1`

`\cos x=\pm1`

`x=0\:or\:x=\pi`

if the interval is
`0\le x\le \pi`, then the two graphs intersect at
`x=0\:or\:x=\pi`

But
`x=0\:and\:x=\pi` does not belong to the open interval
`0\:<\:x\:<\:\pi`

No point of intersection.