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Math Help - Solving for x in the given interval NEED HELP !!!

  1. #1
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    Solving for x in the given interval NEED HELP !!!

    im stuck on three questions and i really need to solve them
    a) sec2x + 1/cosx = 0 x is between and equal to 0 and pie
    b)cos^2x+2sinxcosx-sin^2x=0 x is between and equal to 0 and two pie
    c)2tanx = secx x is between and equal to negative pie and two pie
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  2. #2
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    Hello, math71321!

    b)\;\;\cos^2\!x + 2\sin x\cos x-\sin^2\!x\:=\:0 \qquad0 \leq x \leq 2\pi

    We have: . \underbrace{\cos^2\!x - \sin^2\!x}  + \underbrace{2\sin x\cos x} \;=\;0
    . . . . . . . . . . \cos2x \quad\;\;+ \quad\;\;\sin 2x \quad=\;0

    Then: . \sin2x \:=\:-\cos2x \quad\Rightarrow\quad\frac{\sin2x}{\cos2x} \:=\:-1 \quad\Rightarrow\quad \tan2x \:=\:-1

    Hence: . 2x \;=\;\frac{3\pi}{4},\:\frac{7\pi}{4},\:\frac{11\pi  }{4},\:\frac{15\pi}{4}


    Therefore: . \boxed{x \;=\;\frac{3\pi}{8},\:\frac{7\pi}{8},\:\frac{11\pi  }{8},\:\frac{15\pi}{8}}




    c)\;\;2\tan x \:= \:\sec x\qquad -\pi \leq x \leq 2\pi

    Square both sides: . 4\tan^2\!x \:=\:\sec^2\!x
    . . . . . . . . . . . . 4\overbrace{(\sec^2\!x - 1)} \:=\:\sec^2\!x

    which simplifies to: . 3\sec^2\!x \:=\:4\quad\Rightarrow\quad\sec^2\!x \:=\:\frac{4}{3}\quad\Rightarrow\quad \sec x \:=\:\pm\frac{2}{\sqrt{3}}

    . . Therefore: . \boxed{x \;=\;\pm\frac{\pi}{6},\:\pm\frac{5\pi}{6},\:\frac{  7\pi}{6},\:\frac{11\pi}{6}}

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