G
Guest
Guest
I need to find the general solution for this equation:
\(\displaystyle \L
\sin 2x - 1 = \cos 2x\)
\(\displaystyle \L
2\sin x\cos x - 1 = 1 - 2\sin ^2 x \\
2\sin ^2 x + 2\sin x\cos x - 2 = 0 \\\)
\(\displaystyle {\rm divide by 2 and multiply by - 1} \\\)
\(\displaystyle \L
1 - \sin ^2 x - \sin x\cos x = 0 \\
\cos ^2 x - \sin x\cos x = 0 \\
\cos x(\cos x - \sin x) = 0 \\
\cos x = 0,x = \frac{\pi }{2},\frac{{3\pi }}{2} \\
\cos x = \sin x \\\)
\(\displaystyle {\rm divide both sides by cos x} \\\)
\(\displaystyle \L
\tan x = 1,x = \frac{\pi }
4 \\
\\\)
Is the general solution:
\(\displaystyle \L
2\pi n \pm \frac{\pi }{2},n\pi + \frac{\pi }{4}\)??
Thank you for your help
\(\displaystyle \L
\sin 2x - 1 = \cos 2x\)
\(\displaystyle \L
2\sin x\cos x - 1 = 1 - 2\sin ^2 x \\
2\sin ^2 x + 2\sin x\cos x - 2 = 0 \\\)
\(\displaystyle {\rm divide by 2 and multiply by - 1} \\\)
\(\displaystyle \L
1 - \sin ^2 x - \sin x\cos x = 0 \\
\cos ^2 x - \sin x\cos x = 0 \\
\cos x(\cos x - \sin x) = 0 \\
\cos x = 0,x = \frac{\pi }{2},\frac{{3\pi }}{2} \\
\cos x = \sin x \\\)
\(\displaystyle {\rm divide both sides by cos x} \\\)
\(\displaystyle \L
\tan x = 1,x = \frac{\pi }
4 \\
\\\)
Is the general solution:
\(\displaystyle \L
2\pi n \pm \frac{\pi }{2},n\pi + \frac{\pi }{4}\)??
Thank you for your help