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Question Number 146227 by mathdanisur last updated on 12/Jul/21

$${tg}^2\left(x\right)-\frac{1}{cos\left(x\right)}=1\Rightarrow x=?$$

Answered by gsk2684 last updated on 12/Jul/21

$$\frac{\sin {}^{2}x}{\cos {}^{2}x}-\frac{1}{\cos x}=1$$$$\sin {}^{2}x-\cos x=\cos {}^{2}x$$$$2\cos {}^{2}x+\cos x-1=0$$$$(2\cos x-1)(\cos x+1)=0$$$$\cos x=\frac{1}{2}or-1$$$$x=2n\mathrm{\Pi }\pm \frac{\mathrm{\Pi }}{3}or2n\mathrm{\Pi }+\mathrm{\Pi }wheren\in Z$$

Commented by mathdanisur last updated on 12/Jul/21

$$coolSerthanks$$

Answered by qaz last updated on 12/Jul/21

$${\tan }^2\mathrm{x}-\frac{1}{\cos \mathrm{x}}=\sec {}^2\mathrm{x}-1-\sec \mathrm{x}=1$$$$\Rightarrow \sec {}^2\mathrm{x}-\sec \mathrm{x}-2=0$$$$\Rightarrow \cos \mathrm{x}=-1\mathrm{or}\frac{1}{2}$$$$\Rightarrow \mathrm{x}=(2\mathrm{k}+1)\pi \mathrm{or}(2\mathrm{k}\pm \frac{1}{3})\pi .....(\mathrm{k}\in \mathbb{Z})$$

Commented by mathdanisur last updated on 12/Jul/21

$$coolserthanks$$

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