Esercizio 44 
. Risolvere le seguenti equazioni goniometriche
![\[(1)\quad 2\sin^2(x)=\sin(2x),\qquad (2)\quad 2\sin^2(x)-5\sin(x)-3=0.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-3e4f94f0e710109e12063151b2545f56_l3.png "Rendered by QuickLaTeX.com")
1)
Svolgimento. È utile applicare le formule di duplicazione, ovvero
![\[\sin(2x)=2\sin(x)\cos(x)\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-39c904eab9a4aacc9a86d2e0e89e3c12_l3.png "Rendered by QuickLaTeX.com")
e sostituendo otteniamo:
![\[2\sin^2(x)=2\sin(x)\cos(x)\quad \Leftrightarrow\quad \sin^2(x)-\sin(x)\cos(x)=0\quad \Leftrightarrow\quad \sin(x)(\sin(x)-\cos(x))=0,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6d23b0383298e39b1267b885da932517_l3.png "Rendered by QuickLaTeX.com")
da cui
![\[\sin(x)=0\quad \Leftrightarrow \quad x=k\pi,\;k\in\mathbb{Z}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-430b35368f14a75b0c0b445ffe63a06a_l3.png "Rendered by QuickLaTeX.com")
e
![\[\begin{aligned} & \sin(x)-\cos(x)=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\sin(x)-\dfrac{1}{\sqrt{2}}\cos(x)\right)=\sqrt{2}\left( \dfrac{1}{\sqrt{2}}\cos(x)-\dfrac{1}{\sqrt{2}}\sin(x)\right)=\\ &=\sqrt{2}\left( \cos\left(\dfrac{\pi}{4}\right)\cos(x)-\sin\left(\dfrac{\pi}{4}\right)\sin(x)\right)=\sqrt{2}\cos\left(x+\dfrac{\pi}{4}\right)=0\quad \Leftrightarrow\quad x+\dfrac{\pi}{4}=\dfrac{\pi}{2}+k\pi,\,k\in\mathbb{Z}\quad \Leftrightarrow\quad \\ & \Leftrightarrow\quad x=\dfrac{\pi}{4}+k\pi,\,k\in\mathbb{Z}. \end{aligned}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-6cd2888de2d171c272aeac07277c1514_l3.png "Rendered by QuickLaTeX.com")
Si conclude che la soluzione dell’equazione è
![\[\boxcolorato{analisi}{ \mathcal{S}=\left\{x\in\mathbb{R}:\,x=k\pi ,\, x=\dfrac{\pi}{4}+k\pi,\,\,k\in\mathbb{Z} \right\}.}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-a5db2b6e64aa9f3c6b549eb1f35be21b_l3.png "Rendered by QuickLaTeX.com")
2)
Svolgimento. Si pone 
, ottenendo
![\[2t^2-5t-3=0.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-dba28b9ea3662cd5c9106e0e20e60f21_l3.png "Rendered by QuickLaTeX.com")
Calcoliamo il delta, cioè
![\[\Delta=25-4(-3)(2)=25+24=49,\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-fac91eec34da640ca2dbf40f947c8f15_l3.png "Rendered by QuickLaTeX.com")
da cui
![\[t_{1,2}=\dfrac{5\pm\sqrt{49}}{4}=\dfrac{5\pm7}{4},\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-cadc9d5fefcd07b3d5a4da31f4d2597e_l3.png "Rendered by QuickLaTeX.com")
e quindi
![\[t_1=3\quad \vee\quad t_2=-\dfrac{1}{2},\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-13eae5221ca7c19afedfcfa67cbd30eb_l3.png "Rendered by QuickLaTeX.com")
o anche
![\[\sin(x)=3\quad \vee\quad \sin(x)=-\dfrac{1}{2}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-9f18ccebf249765fc08f25460a8e8a4b_l3.png "Rendered by QuickLaTeX.com")
La prima equazione è chiaramente impossibile in quanto
![\[-1\leq\sin(x)\leq1,\,\forall x \in\mathbb{R}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-d51d203f5b3d9de8cb350e8bce2e2285_l3.png "Rendered by QuickLaTeX.com")
La seconda equazione ha soluzione
![\[x=\dfrac{7\pi}{6}+2k\pi\quad \vee\quad x=\dfrac{11\pi}{6}+2k\pi,\;k\in\mathbb{Z}.\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-7922e98d23c1e69413e1c1b0d02609fd_l3.png "Rendered by QuickLaTeX.com")
Si ha dunque
![\[\mathcal{S}=\left\{\dfrac{7\pi}{6}+2k\pi,\; \dfrac{11\pi}{6}+2k\pi,\;k\in\mathbb{Z}\right\}\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-4d11e60181affbb293f11f979861c7c9_l3.png "Rendered by QuickLaTeX.com")
Si conclude che la soluzione dell’equazione è
![\[\boxcolorato{analisi}{ \mathcal{S}=\left\{x \in \mathbb{R}:\,\dfrac{7\pi}{6}+2k\pi,\; \dfrac{11\pi}{6}+2k\pi,\;k\in\mathbb{Z}\right\}. }\]](https://quisirisolve.com/wp-content/ql-cache/quicklatex.com-c2f21a98f864113ab1d48e893509c270_l3.png "Rendered by QuickLaTeX.com")