Question

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

Answer 1

$sin^{2} 2x-cos^{2} 2x=1$

$x = \frac {1}{4}(2\pi n-\pi)$, n ∈ Ƶ

Answer 2

Let's simplify the given trigonometric expression:

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

Using trigonometric identities, we can express $\sin^2$ and $\cos^2$ in terms of each other:

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

Now, substitute this expression back into the original equation:

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

Distribute the negative sign:

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

Combine like terms:

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

Subtract 1 from both sides:

$-2\cos^2(2x) = 0$

Divide both sides by -2:

$\cos^2(2x) = 0$

Now, take the square root of both sides:

$\cos(2x) = 0$

Now, solve for $x$. The general solution for $\cos(\theta) = 0$ is $\theta = \frac{\pi}{2} + n\pi$, where $n$ is an integer.

So, in this case:

$2x = \frac{\pi}{2} + n\pi$

Now, solve for $x$:

$x = \frac{\pi}{4} + \frac{n\pi}{2}$

So, the solutions to the equation $\sin^2(2x) - \cos^2(2x) = 1$ are $x = \frac{\pi}{4} + \frac{n\pi}{2}$, where $n$ is an integer.