Question

Solve the following trigonometric function on the interval [ 0 , 6.28 ]. sin²x-cos²x=0

Answer 1

`\begin{gathered} \cos ^2x-sin^2x\text{ = }\cos 2x \\ So,\text{ }\sin ^2x-\text{ }\cos ^2x\text{ = -}\cos 2x \end{gathered}`
`\begin{gathered} \text{Thus, -}\cos 2x\text{= 0} \\ In\text{ general }\cos (y)\text{ = 0 implies y = }(n\pi)/(2) \end{gathered}`
`\begin{gathered} -\cos (2x)\text{ =0} \\ 2x\text{ = }(n\pi)/(2) \\ x\text{ =}(n\pi)/(4) \\ \end{gathered}`

The interval [0, 6.28] = [0, 2 x 3.14] = [0 , 2pi]

`\begin{gathered} So\text{ if the interval is (0, 2}\pi) \\ \text{the values are } \\ (\pi)/(4),\text{ }(3\pi)/(4),\text{ }(5\pi)/(4)\text{ , }(7\pi)/(4) \end{gathered}`