solve the equation for the interval 0, 2π).\n\\( \\sin ^ { 2 } ( x ) - \\cos ^ { 2 } ( x ) = 0 \\)\n\\(…

Explanation:

Step1: Use trigonometric identity

We know that (\sin^{2}(x)-\cos^{2}(x)=-\cos(2x)) (double - angle formula (\cos(2x)=\cos^{2}(x)-\sin^{2}(x))). So the equation (\sin^{2}(x)-\cos^{2}(x) = 0) becomes (-\cos(2x)=0), which is equivalent to (\cos(2x)=0).

Step2: Solve for (2x)

If (\cos\theta = 0), then (\theta=\frac{\pi}{2}+k\pi), (k\in\mathbb{Z}). Let (\theta = 2x), so (2x=\frac{\pi}{2}+k\pi).

Step3: Solve for (x)

Divide both sides of (2x=\frac{\pi}{2}+k\pi) by (2), we get (x=\frac{\pi}{4}+\frac{k\pi}{2}).

Step4: Find solutions in the interval ([0,2\pi))

When (k = 0), (x=\frac{\pi}{4}); when (k = 1), (x=\frac{\pi}{4}+\frac{\pi}{2}=\frac{3\pi}{4}); when (k = 2), (x=\frac{\pi}{4}+\pi=\frac{5\pi}{4}); when (k = 3), (x=\frac{\pi}{4}+\frac{3\pi}{2}=\frac{7\pi}{4}).

Answer:

A. (\left{\frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}\right})