Bu masala bo'yicha maslahatda ta'kidlanganidek, ayiruvchi ham, maxraj ham uzluksiz bo'lgan ratsional ifodani ko'rib chiqishda (bizda bu erda bo'lgani kabi, chunki ayirma doimiy va maxraj uzluksiz funktsiyalar yig'indisidir) ratsional ifoda uzluksiz bo'lgan yagona nuqta nolga bo'linish bo'ladi.

Shuning uchun biz qilishimiz kerak bo'lgan yagona narsa maxrajning nolga teng ekanligini aniqlashdir. Agar trig funksiyalari ishtirokidagi tenglamalarni qanday yechish kerakligini eslay olmasangiz, “Ko‘rib chiqish” bo‘limiga qaytib, u yerdagi Trig tenglamalarini yechish bo‘limlarini ko‘rib chiqing.

Mana, maxrajning nolga tengligini aniqlash uchun yechim ish. Kalkulyatorimiz yordamida biz quyidagilarni olamiz:

\[2 - 4\cos \left( {3z} \right) = 0\hspace{0.5in} \to \hspace{0.5in}\,\,3z = {\cos ^{ - 1}}\left( {\frac{1}{2}} \right) = 1.0472\]

Ikkinchi burchak to'rtinchi kvadrantda bo'ladi va \(2\pi - 1.0472 = 5.2360\).

Shunday qilib, maxraj nolga teng bo'ladi,

\[\begin{array}{rl}{3z = 1.0472 + 2\pi n}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{3z = 5.2360 + 2\pi n}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\\{z = 0.3491 + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = 1.7453 + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

Shuning uchun funksiya nuqtalarda uzilib qoladi,

\[\begin{array}{rl}{z = 0.3491 + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = 1.7453 + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]

Shuni ham yodda tutingki, bu sizning asosiy birlik doirangiz qiymatlarini bilsangiz, echilishi mumkin bo'lgan bir nechta trig tenglamalardan biri edi. Mana uzilish nuqtalari uchun aniq yechim.

\[\begin{array}{rl}{z = \frac{\pi }{9} + \frac{{2\pi n}}{3}}{\hspace{0.25in}{\mbox{OR}}\hspace{0.25in}}{z = \frac{{5\pi }}{9} + \frac{{2\pi n}}{3}}{\,\,\,\,\,n = 0, \pm 1, \pm 2, \ldots }\end{array}\]