Question

Sin^2(x)=cos^2(x). Solve for the equation for the interval [0, 2pi)

Answer 1

`\sin^2x=\cos^2x\ \ \ \ |we\ know:\sin^2x+\cos^2x=1\to\cos^2x=1-\sin^2x\\\\\sin^2x=1-\sin^2x\ \ \ \ \ |add\ \sin^2x\ to\ both\ sides\\\\2\sin^2x=1\ \ \ \ |divide\ both\ sides\ by\ 2\\\\\sin^2x=(1)/(2)\iff\sin x=-\sqrt{(1)/(2)}\ or\ \sin x=\sqrt{(1)/(2)}\\\\x=-(\sqrt2)/(2)\ or\ x=(\sqrt2)/(2)\\\\Answer:\ \boxed{x=(\pi)/(4)\ or\ x=(3\pi)/(4)\ or\ x=(5\pi)/(4)\ or\ x=(7\pi)/(4)}`