Question

Find all solutions in the interval [0, 2π). tan x + sec x = 1

Answer 1

`\bf sin^2(\theta)+cos^2(\theta)=1\implies cos^2(\theta)=1-sin^2(\theta) \\\\\\ cos(\theta)=√(1-sin^2(\theta)) \\\\\\ and\qquad tan(\theta)=\cfrac{sin(\theta)}{cos(\theta)}\qquad sec(\theta)=\cfrac{1}{cos(\theta)}\\\\ -------------------------------\\\\`


`\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{√(1-sin^2(x))} \\\\\\\ [sin(x)+1]^2=[√(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=(3\pi )/(2) \end{cases}`