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Math Help - solve

  1. #1
    Newbie
    Joined
    Dec 2007
    Posts
    3

    solve

    A.2sinx+1=0
    B.2cos^2x-1=0
    C.3tan^3x-tanx=0
    D.2sin^2x=2+cosx

    I really need help.. I have my answers here but im not sure if they are correct:

    A.x=7/6pi, 11/6pi
    B.x=pi/4 7/4pi
    C.x=pi/6 7/6pi 2pi pi
    D.x=pi/2, 3/2pi pi/3 5/3pi


    I also need these checked
    Using a sum of difference identity find exact values for the following:

    A.Cos75(degrees)
    B.Sin15(degrees)
    C.tan345(degrees)

    I got
    A.*squareroot*6-2/4
    B.*squareroot*6-2/4
    C.3-*squareroot*3/3

    For problems 2 and 3 find the following.
    A.cos(x-y): sin(x-y); tan (x-y); the quadrant (x-y) is in.
    B.cos(x+y); sin(x+y) tan(x+y) the quadrant (x+y) is in.

    2.If (pi/2)<x<pi and pi<y<(3pi/2) and cosx=-5/13 and tany=3/4

    3.If pi<x< or equal too(3pi/2) and (3pi/2)<y<2pi, and secx=-41/9 and tany=-12/5

    If you could show me how to do them it would be appreciated.
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  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, ninelivesmeow!


    I'll run through the first one . . .


    A)\;\;2\sin x+1\:=\:0
    We have: . \sin x \:=\:-\frac{1}{2}\quad\Rightarrow\quad x \:=\:\frac{7\pi}{6},\:\frac{11\pi}{6}

    . . Your answer is correct!



    B)\;\;2\cos^2\!x-1\:=\:0
    You missed a few . . .

    We have: . \cos^2\!x \:=\:\frac{1}{2}\quad\Rightarrow\quad \cos x \:=\:{\color{red}\pm}\frac{1}{\sqrt{2}}

    Therefore: . \boxed{x \;=\;\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{  4},\:\frac{7\pi}{4}}



    C)\;\;3\tan^3\!x-\tan x\:=\:0

    We have: . \tan x(3\tan^2\!x - 1) \:=\:0

    Then: . \tan x \:=\:0\quad\Rightarrow\quad\boxed{ x \:=\:0,\:\pi,\:2\pi}

    And: . \tan^2\!x \:=\: \frac{1}{3} \quad\Rightarrow\quad \tan x \:=\: {\color{red}\pm}\frac{1}{\sqrt{3}}\quad\Rightarrow  \quad \boxed{x \;=\;\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{7\pi}{  6},\:\frac{11\pi}{6}}



    D)\;\;2\sin^2\!x\:=\:2+\cos x
    We have: . 2(1-\cos^2\!x) \:=\:2+\cos x

    . . which simplifies to: . 2\cos^2\!x + \cos x \:=\:0\quad\Rightarrow\quad \cos x(2\cos x + 1) \:=\:0


    Then: . \cos x \:=\:0\quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}

    And: . 2\cos x + 1 \:=\:0\quad\Rightarrow\quad \cos x \:=\:-\frac{1}{2}\quad\Rightarrow\quad\boxed{ x \;=\;\frac{2\pi}{3},\:\frac{4\pi}{3}}

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