Mathematics, 28.01.2020 08:31, lillianbrowning10
Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x
Answers: 2
![\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\sqrt{1} \\\\\\ sin(x)=\pm 1\implies \measuredangle x=sin^{-1}(\pm1)\implies \measuredangle x= \begin{cases} \frac{\pi }{2}\\\\ \frac{3\pi }{2} \end{cases}](/tpl/images/0476/0485/0c430.png)
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Find all solutions in the interval [0,2pi). sec^2x-2=tan^2x...
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