Question

Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x

Answer 1

`\bf tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad \qquad sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta)\\\\ -----------------------------\\\\ sec^2(x)-2=tan^2(x)\qquad [0,2\pi ) \\\\\\ \cfrac{1^2}{cos^2(x)}-2=\cfrac{sin^2(x)}{cos^2(x)}\implies \cfrac{1}{cos^2(x)}-2-\cfrac{sin^2(x)}{cos^2(x)}=0`


`\bf \cfrac{1-2cos^2(x)-sin^2(x)}{cos^2(x)}=0\implies 1-2\underline{cos^2(x)}-sin^2(x) \\\\\\ 1-2\underline{[1-sin^2(x)]}-sin^2(x)=0\implies 1-2+2sin^2(x)-sin^2(x)=0 \\\\\\ -1+sin^2(x)=0\implies sin^2(x)=1\implies sin(x)=\pm√(1) \\\\\\ sin(x)=\pm 1\implies \measuredangle x=sin^(-1)(\pm1)\implies \measuredangle x= \begin{cases} (\pi )/(2)\\\\ (3\pi )/(2) \end{cases}`