[tex3]
3\sin^2(e^x) - 2 \sqrt3.\sin(e^x).\cos(e^x) - 3\cos^2(e^x) = 0\quad\textbf{(1)}\\[24pt]
\text{Seja }z=e^{ie^x}: \sin(e^x)=\dfrac{z-\overline{z}}{2i}\text{ e }\cos(e^x)=\dfrac{z+\overline{z}}{2}\\
\text{e }|z|=1\text{ ou seja }\mathrm{Re}^2(z)+\mathrm{Im}^2(z)=1\\[24pt]
\begin{align}(1)&\iff \frac{3}{-4}(z-\overline{z})^2-\frac{2\sqrt{3}}{4i}(z+\overline{z})(z-\overline{z})-\frac{3}{4}(z+\overline{z})^2=0\\[12pt]
&\iff -\frac{3}{4}(i2\mathrm{Im}(z))^2-\frac{2\sqrt{3}}{4i}\cdot2\mathrm{Re}(z)\cdot i2\mathrm{Im}(z)-\frac{3}{4}(2\mathrm{Re}(z))^2=0\\[12pt]
&\iff 3\mathrm{Im}^2(z)-2\sqrt{3}\mathrm{Re}(z)\mathrm{Im}(z)-3\mathrm{Re}^2(z)=0\\[12pt]
&\iff3\mathrm{Im}^2(z)-2\sqrt{3}\sqrt{1-\mathrm{Im}^2(z)}\ \mathrm{Im}(z)-3(1-\mathrm{Im}^2(z))=0\\[12pt]
&\iff6\mathrm{Im}^2(z)-3=2\sqrt{3}\mathrm{Im}(z)\sqrt{1-\mathrm{Im}^2(z)}\\[12pt]
&\iff 16\mathrm{Im}^4(z)-16\mathrm{Im}^2(z)+3=0\\[12pt]
&\iff \left\{\begin{array}{}\mathrm{Im}^2(z)=\dfrac{3}{4}\\\text{ou}\\\mathrm{Im}^2(z)=\dfrac{1}{4}\end{array}\right.\\[12pt]
&\iff \left\{\begin{array}{}|\mathrm{Im}(z)|=\dfrac{\sqrt{3}}{2}\text{ e }|\mathrm{Re}(z)|=\dfrac{1}{2}\\\text{ou}\\|\mathrm{Im}(z)|=\dfrac{1}{2}\text{ e }|\mathrm{Re}(z)|=\dfrac{\sqrt{3}}{2}\end{array}\right.\\[12pt]
&\iff \left\{\begin{array}{}z=e^{i(\frac{\pi}{3}+k\pi)}\\\text{ou}\\z=e^{i(\frac{2\pi}{3}+k\pi)}\\\text{ou}\\z=e^{i(\frac{\pi}{6}+k\pi)}\\\text{ou}\\z=e^{i(\frac{5\pi}{6}+k\pi)}\end{array}\right.\quad k\in\mathbb{Z}\\[12pt]
&\iff \left\{\begin{array}{}e^x=\frac{\pi}{3}+k\pi\\\text{ou}\\e^x=\frac{2\pi}{3}+k\pi\\\text{ou}\\e^x=\frac{\pi}{6}+k\pi\\\text{ou}\\e^x=\frac{5\pi}{6}+k\pi\end{array}\right.\quad k\in\mathbb{N}\text{ já que }e^x>0\\[12pt]
&\iff e^x=k\frac{\pi}{6},\quad k\in\mathbb{N}^*\text{ e }k\neq 3p,\ p\in\mathbb{N}\quad\small\text{todos os múltiplos de }\frac{\pi}{6}\text{ menos os múltiplos de }\frac{\pi}{2}\\
&\iff x=\ln(k\frac{\pi}{6}),\quad k\in\mathbb{N}^*\text{ e }k\neq 3p,\ p\in\mathbb{N}\\
\end{align}
[tex3]
3\sin^2(e^x) - 2 \sqrt3.\sin(e^x).\cos(e^x) - 3\cos^2(e^x) = 0\quad\textbf{(1)}\\[24pt]
\text{Seja }z=e^{ie^x}: \sin(e^x)=\dfrac{z-\overline{z}}{2i}\text{ e }\cos(e^x)=\dfrac{z+\overline{z}}{2}\\
\text{e }|z|=1\text{ ou seja }\mathrm{Re}^2(z)+\mathrm{Im}^2(z)=1\\[24pt]
\begin{align}(1)&\iff \frac{3}{-4}(z-\overline{z})^2-\frac{2\sqrt{3}}{4i}(z+\overline{z})(z-\overline{z})-\frac{3}{4}(z+\overline{z})^2=0\\[12pt]
&\iff -\frac{3}{4}(i2\mathrm{Im}(z))^2-\frac{2\sqrt{3}}{4i}\cdot2\mathrm{Re}(z)\cdot i2\mathrm{Im}(z)-\frac{3}{4}(2\mathrm{Re}(z))^2=0\\[12pt]
&\iff 3\mathrm{Im}^2(z)-2\sqrt{3}\mathrm{Re}(z)\mathrm{Im}(z)-3\mathrm{Re}^2(z)=0\\[12pt]
&\iff3\mathrm{Im}^2(z)-2\sqrt{3}\sqrt{1-\mathrm{Im}^2(z)}\ \mathrm{Im}(z)-3(1-\mathrm{Im}^2(z))=0\\[12pt]
&\iff6\mathrm{Im}^2(z)-3=2\sqrt{3}\mathrm{Im}(z)\sqrt{1-\mathrm{Im}^2(z)}\\[12pt]
&\iff 16\mathrm{Im}^4(z)-16\mathrm{Im}^2(z)+3=0\\[12pt]
&\iff \left\{\begin{array}{}\mathrm{Im}^2(z)=\dfrac{3}{4}\\\text{ou}\\\mathrm{Im}^2(z)=\dfrac{1}{4}\end{array}\right.\\[12pt]
&\iff \left\{\begin{array}{}|\mathrm{Im}(z)|=\dfrac{\sqrt{3}}{2}\text{ e }|\mathrm{Re}(z)|=\dfrac{1}{2}\\\text{ou}\\|\mathrm{Im}(z)|=\dfrac{1}{2}\text{ e }|\mathrm{Re}(z)|=\dfrac{\sqrt{3}}{2}\end{array}\right.\\[12pt]
&\iff \left\{\begin{array}{}z=e^{i(\frac{\pi}{3}+k\pi)}\\\text{ou}\\z=e^{i(\frac{2\pi}{3}+k\pi)}\\\text{ou}\\z=e^{i(\frac{\pi}{6}+k\pi)}\\\text{ou}\\z=e^{i(\frac{5\pi}{6}+k\pi)}\end{array}\right.\quad k\in\mathbb{Z}\\[12pt]
&\iff \left\{\begin{array}{}e^x=\frac{\pi}{3}+k\pi\\\text{ou}\\e^x=\frac{2\pi}{3}+k\pi\\\text{ou}\\e^x=\frac{\pi}{6}+k\pi\\\text{ou}\\e^x=\frac{5\pi}{6}+k\pi\end{array}\right.\quad k\in\mathbb{N}\text{ já que }e^x>0\\[12pt]
&\iff e^x=k\frac{\pi}{6},\quad k\in\mathbb{N}^*\text{ e }k\neq 3p,\ p\in\mathbb{N}\quad\small\text{todos os múltiplos de }\frac{\pi}{6}\text{ menos os múltiplos de }\frac{\pi}{2}\\
&\iff x=\ln(k\frac{\pi}{6}),\quad k\in\mathbb{N}^*\text{ e }k\neq 3p,\ p\in\mathbb{N}\\
\end{align}
O conjunto solução de (tg^2x – 1)(1 – cotg^2x) = 4,
x ≠ \frac{kπ}{2}, k ∈ Z, é
a) \frac{π}{3} + \frac{kπ}{3} , k \in Z
b) \frac{π}{4} + \frac{kπ}{4} , k \in Z
c) \frac{π}{6} + \frac{kπ}{4} , k \in...