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Math Help - Trig Help Please!!!

  1. #1
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    Trig Help Please!!!

    I don't know what to do. Can someone please help me solve these. I don't even know how to start.


    1. Find all solutions in the interval [0, 2pie) of 2cos(x)-square root 3=0

    2. Find all solutions in the interval [0, 2pie) of sin(x)=(1/4sin(x))

    3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)

    4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0
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  2. #2
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    Quote Originally Posted by jrose011 View Post
    I don't know what to do. Can someone please help me solve these. I don't even know how to start.


    1. Find all solutions in the interval [0, 2pie) of 2cos(x)-square root 3=0

    2. Find all solutions in the interval [0, 2pie) of sin(x)=(1/4sin(x))

    3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)

    4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0

    (1) 2cosx-\sqrt{3}=0

    cosx=\frac{\sqrt{3}{2}}

    x=30 , 330

    (2) sinx = \frac{1}{4sinx }

     <br />
4sin^2x=1<br />
    sinx =1/2 or sinx =-1/2
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  3. #3
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    Hello jrose011
    Quote Originally Posted by jrose011 View Post
    ...

    3. Find all solutions in the interval [0, 2pie) of 3sec^2(x)tan(x)=4tan(x)
    With an equation like

    3\sec^2x\tan x = 4\tan x

    you have to be careful that you don't miss any solutions by dividing both sides by \tan x, to get:

    3\sec^2x = 4

    Remember the first commandment of arithmetic, which is "Thou shalt not divide by zero."

    You see, \tan x = 0 is a possible solution, isn't it? (Because 3\sec^2x \times 0 = 0 = 4 \times 0.) And if you divide both sides by it, you'll lose this solution.

    So you have to say:

    3\sec^2x\tan x = 4\tan x

    \Rightarrow \tan x = 0 or 3\sec^2 x = 4

    So you now have 2 equations to solve. \tan x = 0 is very easy. Can you do the second one? Write it as:

    \frac{3}{\cos^2x} = 4

    \Rightarrow \cos^2x = \frac{3}{4}

    ... and then don't forget the second commandment of arithmetic "When thou takest square roots of both sides, remember thy plus-or-minus sign".

    4. Find all solutions in the interval [0, 2pie) of sin[x+(pie/4)]+sin[x-(pie/4)]=0
    Use \sin(A+B) = \sin A \cos B + \cos A \sin B and \sin(A-B) = \sin A \cos B - \cos A \sin B:

    \sin(x + \tfrac{\pi}{4}) + \sin(x - \tfrac{\pi}{4}) = 0

    \Rightarrow \sin x \cos\tfrac{\pi}{4} + \cos x \sin\tfrac{\pi}{4} +\sin x \cos\tfrac{\pi}{4} - \cos x \sin\tfrac{\pi}{4} =0

    \Rightarrow 2\sin x \cos\tfrac{\pi}{4} = 0

    \Rightarrow \sin x = 0

    \Rightarrow x = \dots

    Grandad
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