Respuesta :
We have to solve the following
[tex]3\sin (2x)=-2\cos ^2(2x)[/tex]We need to find a trigonometric identity
[tex]\cos ^2(2x)=1-\sin ^2(2x)[/tex]then
[tex]\begin{gathered} 3\sin (2x)=-2\cdot(1-\sin ^2(2x)) \\ 3\sin (2x)=-2+2\sin ^2(2x) \\ \end{gathered}[/tex]we can rewrite this as
[tex]2\sin ^2(2x)-3\sin (2x)-2=0[/tex]We can make the following substitution
[tex]\begin{gathered} u=\sin (2x) \\ \Rightarrow2u^2-3u-2=0 \end{gathered}[/tex]Now, solve with the quadratic formula
[tex]\begin{gathered} x_{1,\: 2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \end{gathered}[/tex][tex]u_{1,\: 2}=\frac{-\left(-3\right)\pm\sqrt{\left(-3\right)^2-4\cdot\:2\left(-2\right)}}{2\cdot\:2}[/tex][tex]u_1=\frac{-\left(-3\right)+5}{2\cdot\:2},\: u_2=\frac{-\left(-3\right)-5}{2\cdot\:2}[/tex]So, we get there are two values for u
[tex]u=2,\: u=-\frac{1}{2}[/tex]Now, let's substitute back
[tex]\begin{gathered} u=\sin (2x) \\ \Rightarrow\sin \mleft(2x\mright)=-\frac{1}{2},\: \sin \mleft(2x\mright)=2 \end{gathered}[/tex]we're almost done!
Now, let's find x for each case
[tex]\begin{gathered} \sin \mleft(2x\mright)=-\frac{1}{2} \\ 2x=\frac{7\pi}{6}+2\pi n,\: 2x=\frac{11\pi}{6}+2\pi n \\ x=\frac{7\pi}{12}+\pi n,\: x=\frac{11\pi}{12}+\pi n \end{gathered}[/tex]for the other one there is no solution
[tex]\sin \mleft(2x\mright)=2\quad \colon\quad \mathrm{No\: Solution\: for}\: x\in\mathbb{R}[/tex]So the answer in degrees from 0° to 360° is:
[tex]\begin{gathered} x=105^{\circ}+180^{\circ\: }n,\: x=165^{\circ}+180^{\circ\: }n \\ x=105^{\circ\: },165^{\circ},285^{\circ\: },345^{\circ\: } \end{gathered}[/tex]Answer: 105°, 165°, 285° and 345°
