Question

Si aceste:cos x=2/3,x aparține (pi/2,pi);
ctgx=-4/3,x aparține (pi/2,pi);

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Answer 1

a)

$\displaystyle x\in\left(\frac\pi2,\pi\right)\Rightarrow \sin x>0\text{ \c si }\cos x<0\\\\\displaystyle tg x=-2\Rightarrow \frac{\sin x}{\cos x}=-2\Rightarrow \sin x=-2\cos x\\\sin^2x+\cos^2x=1\Rightarrow (-2\cos x)^2+\cos^2x=1\Rightarrow 5\cos^2x=1\Rightarrow \cos^2x=\frac15\Rightarrow \cos x=\pm\sqrt{\frac15}\\\text{Cum }\cos x<0\Rightarrow \cos x=-\frac1{\sqrt5}=-\frac{\sqrt5}5\\\sin x=\frac{2\sqrt5}{5}\\ctg x=\frac1{tg x}=\frac{-1}2$

b)

$\displaystyle x\in\left(0,\frac\pi2\right)\Rightarrow \sin x>0\text{ \c si }\cos x>0\\\sin x=0.6=\frac{6}{10}=\frac{3}{5}\\\sin^2x+\cos^2x=1\Rightarrow \frac{9}{25}+\cos^2x=1\Rightarrow \cos^2x=\frac{16}{25}\Rightarrow\cos x=\pm\frac45\\\text{Cum }\cos x>0\Rightarrow \cos x=\frac45\\\displaystyle tg x=\frac{\sin x}{\cos x}=\frac{\frac35}{\frac45}=\frac34\\\\ctg x=\frac{1}{tg x}=\frac43$