A' fuasgladh cho-aontaran triantanachd ann an radianan (leudachadh)
Gus co-aontaran triantanachd fhuasgladh far a bheil tomhas radian, bu chòir dhut an aon dòigh a leantainn 's a bh' agad le ceuman. Cleachd an diagram gu h-ìosal gus do chuideachadh a' cleachdadh nan riaghailtean iomchaidh.
Bu chòir cuideachd gum biodh fios agad air na luachan mionaideach bho thriantain nan luachan mionaideach.
Eisimpleir 1
Fuasgail \(2\cos (x) = 1\), airson \(0 \le x \le 2\pi\).
Fuasgladh
\(2\cos (x) = 1\)
\(\cos (x) = \frac{1}{2}\)
\(x = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right)\)
Fuasgail san aon dòigh 's a dhèanadh tu nam b' e ceuman a bhiodh ann.
Mar a dhèanadh tu le co-aontaran ann an ceuman, obraich a-mach na cairtealan iomchaidh agus an uair sin cleachd na riaghailtean iomchaidh.
\(x = \frac{\pi }{3}\)
An dara cairteal
\(x = 2\pi - \frac{\pi }{3}\)
\(x = \frac{{5\pi }}{3}\)
Mar sin \(x = \frac{\pi }{3},\,\frac{{5\pi }}{3}\)
Eisimpleir 2
Fuasgail \(\cos 2x + 3\sin x + 1 = 0\), airson \(0 \textless x \textless 2\pi\).
Fuasgladh
\(\cos 2x + 3\sin x + 1 = 0\)
\(1 - 2{\sin ^2}x + 3\sin x + 1 = 0\)
\(- 2{\sin ^2}x + 3\sin x + 2 = 0\)
\(2{\sin ^2}x - 3\sin x - 2 = 0\)
\((2\sin x + 1)(\sin x - 2) = 0\)
\(2\sin x + 1 = 0\)
\(\sin x = - \frac{1}{2}\)
Bhon a tha sin àicheil, tha sinn san treas agus sa cheathramh cairteal.
\(x = {\sin ^{ - 1}}\left( {\frac{1}{2}} \right)\)
\(x = \frac{\pi }{6}\)
An treas cairteal
\(x = \pi + \frac{\pi }{6}\)
\(x = \frac{{7\pi }}{6}\)
An ceathramh cairteal
\(x = 2\pi - \frac{\pi }{6}\)
\(x = \frac{{11\pi }}{6}\)
\(\sin x - 2 = 0\)
\(\sin x = 2\)
Bhon a tha \(0 \textless x \textless 2\pi\) chan eil fuasgladh ann.
Mar sin \(x = \frac{{7\pi }}{6},\,\frac{{11\pi }}{6}\)
