Question

tg x = — √3\2 x apartine [π\2;π] calculati
sin x (1+| cos x |)

Answer 1

[tex]x\in \left[-\dfrac{\pi}{2},\ \pi \right]\ \ \ \ (1) \\\;\\ \it tg x = - \dfrac{\sqrt3}{2}\Rightarrow (tg x )^2= \left(- \dfrac{\sqrt3}{2}\right)^2\Rightarrow \dfrac{\sin^2 x}{\cos^2 x } = \dfrac{3}{4}[/tex]

Derivam ultima proportie:

$\dfrac{\sin^2 x}{ \cos^2 x +\sin^2 x}=\dfrac{3}{4+3}\Rightarrow \dfrac{\sin^2 x}{1}=\dfrac{3}{7}\Rightarrow \sqrt{\sin^2 x}=\dfrac{\sqrt3}{\sqrt7}$

[tex]\Rightarrow \sin x=\pm \dfrac{\sqrt3}{\sqrt7}\stackrel{(1)}{\Longrightarrow}\sin x= \dfrac{\sqrt21} {7}\ \ \ \ (2) \\\;\\ \cos^2 x + \sin^2 x=1\Rightarrow \cos^2 x +\dfrac{3}{7}=1\Rightarrow \cos^2 x=1-\dfrac{3}{7} \\\;\\ \Rightarrow \cos^2 x=\dfrac{4}{7}\Rightarrow \sqrt{\cos^2 x} =\sqrt{\dfrac{4}{7}}\Rightarrow \cos x =\pm \dfrac{2}{\sqrt 7} \stackrel{(1)}{\Longrightarrow} \\\;\\ \Rightarrow \cos x =- \dfrac{2\sqrt 7}{7}\ \ \ \ (3)[/tex]

[tex]\sin x(1+|\cos x|) \stackrel{(2),(3)}{=} \ \dfrac{\sqrt{21}}{7}(1+|-\dfrac{2\sqrt7}{7}|)= \\\;\\ =\dfrac{\sqrt{21}}{7}(1+\dfrac{2\sqrt7}{7}) = \dfrac{\sqrt21}{7}+\dfrac{\sqrt{21}\cdot2\sqrt7}{7\cdot7}=\dfrac{\sqrt{21}}{7}+\dfrac{\sqrt3\cdot\sqrt7\cdot2\sqrt7}{7\cdot7} \\\;\\ =\dfrac{\sqrt{21}}{7}+\dfrac{2\sqrt3}{7}=\dfrac{\sqrt{21}+2\sqrt3}{7}=\dfrac{\sqrt3(\sqrt7+2)}{7}[/tex]