حل تمرین معادلات مثلثاتی

فصل2: مثلثات، درس2: معادلات مثلثاتی

تمرینات پایانی درس2

1-معادلات زیر را حل کنید.

\[\text{الف})sin\frac{\pi}{2}=sin3x\]

\(3x = 2k\pi+\frac{\pi}{2} \rightarrow x = \frac{2k\pi}{3} + \frac{\pi}{6}\ ,\ 3x = 2k\pi+\pi - \frac{\pi}{2} = 2k\pi+\frac{\pi}{2}\ \rightarrow x = \frac{2k\pi}{3} + \frac{\pi}{6}\)

\[\text{ب})cos 2x-cosx+1=0\]

\(2\cos^{2}x - 1 - cosx + 1 = 0 \rightarrow 2\cos^{2}x - cosx = 0 \rightarrow cosx(2cosx - 1) = 0 \rightarrow    cosx = 0 \rightarrow x = k\pi+\frac{\pi}{2}\ ,\ 2cosx - 1 = 0 \rightarrow cosx = \frac{1}{2} = \cos\frac{\pi}{3} \rightarrow x = 2k\pi \pm \frac{\pi}{3}\)

\[\text{پ})cos x=cos2x\]

\([2x = 2k\pi + x \rightarrow x = 2k\pi\ ,\ 2x = 2k\pi - x \rightarrow 3x = 2k\pi \rightarrow x = \frac{2k\pi}{3}]\)

\[\text{ت})cos 2x-3sinx+1=0\]

\[1 - {2sin}^{2}x - 3sinx + 1 = 0 \rightarrow {2sin}^{2}x + 3sinx - 2 = 0 \rightarrow {2u}^{2} + 3u - 2 = 0 \rightarrow\]

\([\mathrm{\Delta} = 9 - 4(2)( - 2) = 25 \rightarrow u = \frac{-3 \pm \sqrt{25}}{4} = - 2,\frac{1}{2} \rightarrow sinx = - 2ق\ ق\ غ\ ,\ sinx = \frac{1}{2} = \sin\frac{\pi}{6}]\)

\[x = 2k\pi + \frac{\pi}{6}\ ,\ x = 2k\pi + \pi - \frac{\pi}{6} = 2k\pi + \frac{5\pi}{6}\]

\[{ث)cos}^{2}x - sinx = \frac{1}{4}\]

\([{1 - \sin}^{2}x - sinx - \frac{1}{4} = 0 \rightarrow \sin^{2}x + sinx - \frac{3}{4} = 0 \rightarrow u^{2} + u - \frac{3}{4} = 0 \rightarrow \mathrm{\Delta} = 1 - 4(1)\left( - \frac{3}{4} \right) = 4 \rightarrow u = \frac{-1 \pm \sqrt{4}}{2} = - \frac{3}{2}\ ,\frac{1}{2} \rightarrow sinx = - \frac{3}{2}\ ق\ ق\ غ\ ,\ sinx = \frac{1}{2}{= \sin}\frac{\pi}{6} \rightarrow]\)

\([x = 2k\pi + \frac{\pi}{6}\ ,\ x = 2k\pi + \pi - \frac{\pi}{6} = 2k\pi + \frac{5\pi}{6}]\)

\({ج)}^{}sinx - cos2x = 0 \rightarrow cos2x = sinx \rightarrow cos2x = \cos\left( \frac{\pi}{2} - x \right) \rightarrow 2x = 2k\pi \pm \left( \frac{\pi}{2} - x \right)\)

\[\left\{ \begin{array}{r} 2x = 2k\pi + \frac{\pi}{2} - x \rightarrow 3x = 2k\pi + \frac{\pi}{2} \rightarrow x = \frac{2k\pi}{3} + \frac{\pi}{6} \\ 2x = 2k\pi - \frac{\pi}{2} + x \rightarrow x = 2k\pi - \frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \right.\]
 

\[{چ)}^{}\tan(2x - 1) = 0\]

\[\tan(2x - 1) = 0 = tan0 \rightarrow 2x - 1 = k\pi + 0 \rightarrow 2x = k\pi + 1 \rightarrow x = \frac{k\pi}{2} + \frac{1}{2}\]

\[{ح)}^{}\tan{3x} = \tan{\pi x}\]

\[3x = k\pi + \ \pi x \rightarrow x(3 - \pi) = k\pi \rightarrow x = \frac{k\pi}{3 - \pi}\]

2-مثلثی با مساحت 3 سانتی متر مربع مفروض است. اگر اندازه­ی دو ضلع آن 2و6 سانتی متر باشد، چند مثلث با این خاصیت­ها می­توان ساخت.

مساحت مثلث\(S = \frac{1}{2}\ absin\theta \rightarrow 30 = \frac{1}{2} \times 2 \times 6 \times sin\theta \rightarrow sin\theta = \frac{3}{6} = \frac{1}{2} \rightarrow \theta = 30{^\circ},150{^\circ} \rightarrow دومثلث\)