Równanie trygonometryczne

[km1992]

Nie wiem jak rozwiązać równanie: \(\displaystyle{ \tan(\frac{\pi}{4}+x) + \tan(x) -2 = 0}\)

[Pancernik]

\(\displaystyle{ \tg\left(\alpha+\beta\right)=\frac{\tg\alpha+\tg\beta}{1-\tg\alpha\cdot\tg\beta}\\
\frac{\pi}{4}=45^\circ\\
\tg\left(45^\circ+x\right)+\tg x-2=0\\
\frac{\tg 45^\circ+\tg x}{1-\tg 45^\circ\cdot\tg x}+\tg x-2=0\\
\frac{1+\tg x}{1-1\cdot\tg x}+\tg x-2=0\\
\frac{1+\tg x}{1-\tg x}+\tg x-2=0\\
1+\tg x+\tg x\left(1-\tg x\right)-2\left(1-\tg x\right)=0\\
1+\tg x+\tg x-\tg^2x-2+2\tg x=0\\
-\tg^2x+4\tg x-1=0\\
\Delta=16-4=12\\
\tg_1x=\frac{-4-2\sqrt{3}}{-2}=2+\sqrt{3}\\
x_1=75^\circ\\
\tg_2x=\frac{-4+2\sqrt{3}}{-2}=2-\sqrt{3}\\
x_2=15^\circ}\)

Sprawdzenie:
\(\displaystyle{ \tg\left(45^\circ+x\right)+\tg x-2=0\\
1^\circ\quad x=75^\circ\\
\tg\left(45^\circ+75^\circ\right)+\tg 75^\circ-2=0\\
\tg\left(90^\circ+30^\circ\right)+\tg 75^\circ-2=0\\
-\ctg30^\circ+\tg 75^\circ-2=0\\
-\sqrt{3}+2+\sqrt{3}-2=0\\
0=0\\
2^\circ\quad x=15^\circ\\
\tg\left(45^\circ+15^\circ\right)+\tg 15^\circ-2=0\\
\tg60^\circ+\tg 15^\circ-2=0\\
\sqrt{3}+2-\sqrt{3}-2=0\\
0=0}\)