Sin^2x-1=-cos^2x

That is simply the trig identity

sin^2 x + cos^2 x = 1

You can show it easily with a right triangle with 90 deg at C where
a^2 + b^2 = c^2

sin A = a/c
cos A = b/c
so
sin^2 A = a^2/c^2
cos^2 A = b^2/c^2
and we know
c^2 sin^2 A + c^2 cos^2 A = c^2
so
sin^2 A + cos^2 A = 1

To solve the equation sin^2(x) - 1 = -cos^2(x), we can use trigonometric identities. Let's start by manipulating the equation using the Pythagorean identity: sin^2(x) + cos^2(x) = 1.

Rearranging the equation, we have:

sin^2(x) - cos^2(x) = -1

Now, let's substitute sin^2(x) = 1 - cos^2(x) based on the Pythagorean identity:

(1 - cos^2(x)) - cos^2(x) = -1

Expanding the equation:

1 - cos^2(x) - cos^2(x) = -1

Combining like terms:

1 - 2cos^2(x) = -1

Subtracting 1 from both sides:

-2cos^2(x) = -2

Finally, dividing both sides by -2:

cos^2(x) = 1

Taking the square root of both sides:

cos(x) = ±1

So, the solutions for the equation sin^2(x) - 1 = -cos^2(x) are:

cos(x) = 1 and cos(x) = -1.

To find the values of x that satisfy these solutions, we can use the inverse cosine function (or arccosine).

For cos(x) = 1, the angle x can be 0 degrees or any integer multiple of 360 degrees:

x = 0° + 360°n, where n is an integer.

For cos(x) = -1, the angle x can be 180 degrees or any integer multiple of 360 degrees:

x = 180° + 360°n, where n is an integer.

So, the solutions to the equation sin^2(x) - 1 = -cos^2(x) are x = 0° + 360°n and x = 180° + 360°n, where n is an integer.