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Find all solutions to the equation (sinx-cosx)^2=1
4 months ago
Q:
Find all solutions to the equation (sinx-cosx)^2=1
Accepted Solution
A:
Answer:[tex]x=\dfrac{\pi k}{2},\ k\in Z.[/tex]Step-by-step explanation:Simplify the equation [tex](\sin x-\cos x)^2=1:[/tex][tex]\sin^2 x-2\sin x\cos x+\cos ^2x=1,\\ \\ (\sin^2x+\cos^2x)-2\sin x\cos x=1.\\[/tex]Since [tex]\sin^2 x+\cos^2 x=1[/tex]and [tex]2\sin x\cos x=\sin 2x,[/tex]we have[tex]1-\sin 2x=1,\\ \\\sin 2x=0,\\ \\2x=\pi k,\ k\in Z,\\ \\x=\dfrac{\pi k}{2},\ k\in Z.[/tex]