معادلات مثلثاتی

\(2\sin x - \sqrt 2 = 0 \to \sin x = \frac{{\sqrt 2 }}{2}\left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{4}\\\\x = 2(k + 1)\pi - \frac{\pi }{4}\end{array} \right.\)

\(\cos x = \frac{1}{2} \to \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{3}\\\\x = 2k\pi - \frac{\pi }{3}\end{array} \right.\)

\(\sin 3x = 1\left\{ \begin{array}{l}3x = 2k\pi + \frac{\pi }{2} \to x = \frac{{2k\pi }}{3} + \frac{\pi }{6}\\\\3x = (2k + 1)\pi - \frac{\pi }{2} \to x = \frac{{(2k + 1)\pi }}{3} - \frac{\pi }{6}\end{array} \right.\)

\(\begin{array}{*{20}{l}}\begin{array}{l}\tan 4x.\tan 3x = 1 \to \tan 4x = \frac{1}{{\tan 3x}}\\\\ \to \tan 4x = \cot 3x \to \tan \underbrace u_{} = \tan (\underbrace {\frac{\pi }{2} - 3x}_\alpha )\end{array}\\{}\\\begin{array}{l} \to 4x = 4\pi + \frac{\pi }{2} - 3x \to 7x = k\pi + \frac{\pi }{2}\\\\ \to x = \frac{{k\pi }}{7} + \frac{\pi }{{14}}\end{array}\end{array}\)

\(\begin{array}{*{20}{l}}\begin{array}{l}\tan x.\tan 2x = 1 \to \tan 2x = \frac{1}{{\tan x}}\\\\ \to \tan 2x = \cot x \to \tan 2x = \tan (\frac{\pi }{2} - x)\end{array}\\{}\\\begin{array}{l} \to 2x = k\pi + (\frac{\pi }{2} - x)\\\\ \to 3x = k\pi + \frac{\pi }{2} \to x = \frac{{k\pi }}{3} + \frac{\pi }{6}\end{array}\end{array}\)

توجه داشته باشید که این معادله به ازای برخی از مقادیر بدست آمده برقرار نیست. مثلاً به ازای ۱ = k به دست می آید. \(x = \frac{\pi }{3} + \frac{\pi }{6} = \frac{\pi }{2}\) که جواب معادله نیست.