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Thread: Please help

  1. #1
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    Please help

    Hello. I'm having trouble with the 2 last calculus exercises witch I just can't solve. If someone would please tell me how to solve these two:

    1) 1-sinx -cosx +sin2x+cos2x=0

    2)log(2 below) (4^(x+1)+4) * log(2below) (4^(x)+1)=log(1/square root of 2 below) (square root of 1/8)

    Thanks to everyone who can help me!!!
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  2. #2
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    Does anyone have any ideas or tips how to solve these two, I'm really running out of time
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  3. #3
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    Hello, MadMan!

    $\displaystyle 2)\;\;\log_2\left(4^{x+1}+4\right)\cdot \log_2\left(4^x+1\right) \;=\;\log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1} {8}}\right) $

    The right side is: .$\displaystyle \log_{\frac{1}{\sqrt{2}}}\left(\sqrt{\frac{1}{8}}\ right) \:=\:p \quad\Rightarrow\quad \left(\frac{1}{\sqrt{2}}\right)^p \:=\:\frac{1}{\sqrt{8}} \quad\Rightarrow\quad p \:=\:3$

    The left side is: .$\displaystyle \log_2\bigg[4(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;\bigg[\log_2(4) + \log_2(4^x + 1)\bigg]\cdot\log_2(4^x+1)$

    . . $\displaystyle \;=\bigg[2 + \log_2(4^x+1)\bigg]\cdot\log_2(4^x+1) \;=\;2\cdot\log_2(4^x+1) + \bigg[\log_2(4^x + 1)\bigg]^2$


    The equation becomes: .$\displaystyle \bigg[\log_2(4^x+1)\bigg]^2 + 2\!\cdot\!\bigg[\log_2(4^x+1)\bigg]\:=\:3$
    . . . . $\displaystyle \bigg[\log_2(4^x+1)\bigg]^2 + 2\bigg[\log_2(4^x+1)\bigg] - 3 \;=\;0$

    Factor: . $\displaystyle \bigg[\log_2(4^x+1) +3\bigg]\cdot\bigg[\log_2(4^x+1) -1\bigg] \;=\;0$


    $\displaystyle (1)\;\;\log_2(4^x+1) + 3 \:=\:0\quad\Rightarrow\quad \log_2(4^x+1) \:=\:-3 \quad\Rightarrow\quad 4^x + 1 \:=\:2^{-3}$

    . . . . $\displaystyle 4^x + 1 \:=\:\frac{1}{8} \quad\Rightarrow\quad 4^x \:=\:-\frac{7}{8}$ . . . no real roots


    $\displaystyle (2)\;\;\log_2(4^x+1) - 1 \:=\:0 \quad\Rightarrow\quad \log_2(4^x+1) \:=\:1 \quad\Rightarrow\quad 4^x + 1 \:=\:2^1$

    . . . . $\displaystyle 4^x \:=\:1 \quad\Rightarrow\quad\boxed{ x \:=\:0} $

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  4. #4
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    Hello again, MadMan!

    I'll assume the answers are on the interval [$\displaystyle 0,\:2\pi)$


    $\displaystyle 1)\;\;1-\sin x -\cos x +\sin2x+\cos2x\:=\:0$
    Double-angle identities: .$\displaystyle \begin{array}{ccc}\sin2A &=& 2\sin A\cos A \\ \cos2A &=& 2\cos^2\!A - 1 \end{array}$


    The equation becomes: .$\displaystyle 1 - \sin x - \cos x + 2\sin x\cos x + 2\cos^2\!x - 1 \;=\;0$

    . . and we have: .$\displaystyle 2\sin x\cos x + 2\cos^2\!x - \sin x - \cos x \;=\;0$


    Factor: .$\displaystyle 2\cos x(\sin x + \cos x) - (\sin x + \cos x) \;=\;0$

    Factor: .$\displaystyle (\sin x + \cos x)(2\cos x - 1) \;=\;0 $


    $\displaystyle (1)\;\;\sin x + \cos x \:=\:0 \quad\Rightarrow\quad \sin x \:=\:-\cos x \quad\Rightarrow\quad \frac{\sin x}{\cos x} \:=\:-1$

    . . . . $\displaystyle \tan x \:=\:-1 \quad\Rightarrow\quad\boxed{ x \:=\:\frac{3\pi}{4},\:\frac{7\pi}{4}}$


    $\displaystyle (2)\;\;2\cos x - 1 \:=\:0\quad\Rightarrow\quad \cos x \:=\:\frac{1}{2} \quad\Rightarrow\quad\boxed{ x \:=\:\frac{\pi}{3},\:\frac{5\pi}{3}}$

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  5. #5
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    Soroban, thank you so much, you just made my day!! After being absent from school because of an illness I fell behind and I absolutely could not figure those two out by myself, so A really big thanks to you for helping me out!!
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