Mathematics, 13.12.2019 13:31 jessicawolfking
Find all solutions in the interval [0, 2π). tan x + sec x = 1
Answers: 1
![\bf tan(x)+sec(x)=1\implies \cfrac{sin(x)}{cos(x)}+\cfrac{1}{cos(x)}=1\implies \cfrac{sin(x)+1}{cos(x)}=1 \\\\\\ sin(x)+1=cos(x)\implies sin(x)+1=\boxed{\sqrt{1-sin^2(x)}} \\\\\\\ [sin(x)+1]^2=[\sqrt{1-sin^2(x)}]^2 \\\\\\ sin^2(x)+2sin(x)+1^2=1-sin^2(x) \\\\\\ 2sin^2(x)+2sin(x)=0\implies 2sin(x)[sin(x)+1]=0 \\\\\\ \begin{cases} 2sin(x)=0\\ \qquad sin(x)=0\\ \qquad \measuredangle x=0~,~\pi \\ sin(x)+1=0\\ \qquad sin(x)=-1\\ \qquad \measuredangle x=\frac{3\pi }{2} \end{cases}](/tpl/images/0417/0751/4c40e.png)
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Find all solutions in the interval [0, 2π). tan x + sec x = 1...
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