شاه کلید نکات
شاه کلید سوال
1 معادلات زیر را حل کنید.
الف \(\sin \frac{\pi }{2} = \sin 3x\)
ب cos2x – cosx + 1 = 0
پ cosx = cos2x
ت cos2x – 3sinx + 1 = 0
ث \({\cos ^2}x - \sin x = \frac{1}{4}\)
ج sinx - cos2x = 0
چ tan(2x-1) =0
چ \(\tan 3x = \tan \pi x\)
الف
\(\begin{array}{l} \Rightarrow 3x = 2k\pi + \frac{\pi }{2}\,\,\quad ,\quad k \in \mathbb{Z}\,\\\\ \Rightarrow x = \frac{{2k\pi }}{3} + \frac{\pi }{6}\,\,\quad ,\quad k \in \mathbb{Z}\end{array}\)
ب
\(\begin{array}{l} \Rightarrow (2{\cos ^2}x - 1) - \cos x + 1 = 0\\\\ \Rightarrow 2{\cos ^2} - \cos x = 0\\\\ \Rightarrow \cos x(2\cos x - 1) = 0\\\\(k \in \mathbb{Z})\\\\ \Rightarrow \left\{ \begin{array}{l}\cos x = 0\quad \Rightarrow \quad x = k\pi + \frac{\pi }{2}\\\\\cos x = \frac{1}{2}\quad \Rightarrow \quad x = 2k\pi \pm \frac{\pi }{3}\end{array} \right.\end{array}\)
پ
\(\begin{array}{l} \Rightarrow x = 2k\pi \pm 2x\\\\(k \in \mathbb{Z})\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + 2x \Rightarrow x = - 2k\pi \\\\x = 2k\pi - 2x \Rightarrow x = \frac{{2k\pi }}{3}\end{array} \right.\end{array}\)
ت
\(\begin{array}{l} \Rightarrow - 2{\sin ^2}x + 1 - 3\sin x + 1 = 0\\\\ \Rightarrow - 2{\sin ^2}x - 3\sin x + 2 = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = 25\\\\ \Rightarrow \sin x = \frac{{3 \pm \sqrt {25} }}{{ - 4}}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{3 + 5}}{{ - 4}} = - 2\quad X\\\sin x = \frac{{3 - 5}}{{ - 4}} = - \frac{1}{2}\end{array} \right.\\\\ \Rightarrow \sin x = - \frac{1}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi - \frac{\pi }{6}\\\\x = 2k\pi + \pi + \frac{\pi }{6}\end{array} \right.\end{array}\)
ث
\(\begin{array}{l} \Rightarrow 1 - {\sin ^2}x - \sin x = \frac{1}{4}\\\\ \Rightarrow {\sin ^2}x + \sin x - \frac{3}{4} = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = 4\\\\ \Rightarrow \sin x = \frac{{ - 1 \pm \sqrt 4 }}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}\sin x = \frac{{ - 1 + 2}}{2} = \frac{1}{2}\\\\\sin x = \frac{{ - 1 - 2}}{2} = - \frac{3}{2}\quad X\end{array} \right.\\\\ \Rightarrow \sin x = \frac{1}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\x = 2k\pi + \pi - \frac{\pi }{6}\end{array} \right.\end{array}\)
ج
\(\begin{array}{l} \Rightarrow \sin x + 2{\sin ^2}x - 1 = 0\\\\ \Rightarrow \Delta = {b^2} - 4ac = 9\\\\ \Rightarrow \sin x = \frac{{ - 1 \pm \sqrt 9 }}{4}\\\\ \Rightarrow \sin x = \frac{{ - 1 + 3}}{4} = \frac{1}{2}\\\\ \Rightarrow \left\{ \begin{array}{l}x = 2k\pi + \frac{\pi }{6}\\\\x = 2k\pi + \pi - \frac{\pi }{6}\end{array} \right.\\\\ \Rightarrow \sin x = \frac{{ - 1 - 3}}{4} = - 1\\\\ \Rightarrow x = 2k\pi + \frac{{3\pi }}{2}\end{array}\)
چ
\(\begin{array}{l} \Rightarrow 2x - 1 = k\pi \\\\ \Rightarrow x = \frac{{k\pi + 1}}{2}\end{array}\)
چ
\(\begin{array}{l} \Rightarrow 3x = k\pi + \pi x\\\\ \Rightarrow (3 - \pi )x = k\pi \\\\ \Rightarrow x = \frac{{k\pi }}{{3 - \pi }}\end{array}\)
2 مثلثی با مساحت 3 سانتی متر مربع مفروض است. اگر اندازه دو ضلع آن به ترتیب 2 و 6 سانتی متر باشند، آنگاه چند مثلث با این خاصیت ها می توان ساخت؟
\(\begin{array}{l}S = \frac{1}{2} \times 2 \times 6 \times \sin \alpha = 3\\\\ \Rightarrow \sin \alpha = \frac{1}{2} = \sin \frac{\pi }{6}\\\\ \Rightarrow \alpha = \frac{\pi }{6}\quad ,\quad \frac{{5\pi }}{6}\end{array}\)
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