Trigonometria

por futuromilitar2 Seg 17 Jul 2017, 16:00

Resolver :

$\frac{1}{\sin^2x}=1- \frac{\cos x}{\sin x}$

Resposta:

$x=\frac{\pi}{2}+k\pi \ ou \ x=\frac{3\pi}{4}+k\pi$

por **Victor011** Seg 17 Jul 2017, 16:23

\\\frac{1}{sen^2x}=1-\frac{cosx}{senx}\\\\\frac{sen^2x+cos^2x}{sen^2x}=1-\frac{cosx}{senx}\\\\1+\frac{cos^2x}{sen^2x}=1-\frac{cosx}{senx}\\\\\frac{1}{tan^2x}=-\frac{1}{tanx}\\\\tan^2x=-tanx\;\to\;tanx=0\;ou\;tanx=-1\\\\\bullet\;tanx=0:\\\\\boxed{x=\frac{\pi}{2}+k\pi\;\;,com\;k\in\mathbb{Z}}\\\\\bullet\;tanx=-1:\\\\\boxed{x=\frac{3\pi}{4}+k\pi\;\;,com\;k\in\mathbb{Z}}

Trigonometria

Trigonometria

Re: Trigonometria

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