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Thread: Solve 7 sin 2x - cos 2x = 0

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    Solve 7 sin 2x - cos 2x = 0

    The equation in the title should be 7 sin 2x - cos 2x = 1 not 7 sin 2x - cos 2x = 0. I want to solve 7 sin 2x - cos 2x = 1 for 0\leq{x}\leq\pi. After a few steps I get cos x=0 or 7sinx-cosx=0. From equation 7sinx-cosx=0, at first I assume that sin{x}\neq{0} and I found that x=tan^{-1}7 is one solution. From equation 7sinx-cosx=0, in what way can I proceed for the case sin{x} = 0?
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    Re: Solve 7 sin 2x - cos 2x = 0

    7 sin(2x) - cos(2x) = 1

    $14\sin{x}\cos{x} - (\cos^2{x} - \sin^2{x}) = \cos^2{x} + \sin^2{x}$

    $14\sin{x}\cos{x} - 2\cos^2{x} = 0$

    $\cos{x}(14\sin{x} - 2\cos{x}) = 0$

    $\cos{x} = 0$

    $\tan{x} = \dfrac{1}{7}$

    for $0 \le x < \pi$ ...

    $x = \dfrac{\pi}{2}$

    $x = \arctan\left(\dfrac{1}{7}\right)$
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    Re: Solve 7 sin 2x - cos 2x = 0

    Quote Originally Posted by skeeter View Post
    7 sin(2x) - cos(2x) = 1

    $14\sin{x}\cos{x} - (\cos^2{x} - \sin^2{x}) = \cos^2{x} + \sin^2{x}$

    $14\sin{x}\cos{x} - 2\cos^2{x} = 0$

    $\cos{x}(14\sin{x} - 2\cos{x}) = 0$

    $\cos{x} = 0$

    $\tan{x} = \dfrac{1}{7}$

    for $0 \le x < \pi$ ...

    $x = \dfrac{\pi}{2}$

    $x = \arctan\left(\dfrac{1}{7}\right)$
    From $\cos{x}(14\sin{x} - 2\cos{x}) = 0$, we get $\cos{x}=0$ or $14\sin{x}-2\cos{x}=0$. From equation $14\sin{x}-2\cos{x}=0$, by assuming that $sin{x}\neq{0}$, we get the solution $x=\tan^{-1}(1/7)$. However, for the equation $14\sin{x}-2\cos{x}=0$, we have not yet rule out the possibility that $sin{x}=0$ since we previously assume that $sin{x}\neq{0}$. How can we show that x for which $\sin{x}=0$ is not a solution of $14\sin{x}-2\cos{x}=0$ and hence not a solution of the original equation?
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    Re: Solve 7 sin 2x - cos 2x = 0

    Quote Originally Posted by woo View Post
    How can we show that x for which $\sin{x}=0$ is not a solution of $14\sin{x}-2\cos{x}=0$ and hence not a solution of the original equation?
    $\sin{x} = 0 \implies \cos{x} = \pm 1$

    therefore ...

    $14\sin{x}-2\cos{x}= 14(0) - 2(\pm 1) \ne 0$
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    Re: Solve 7 sin 2x - cos 2x = 0

    Quote Originally Posted by woo View Post
    From $\cos{x}(14\sin{x} - 2\cos{x}) = 0$, we get $\cos{x}=0$ or $14\sin{x}-2\cos{x}=0$. From equation $14\sin{x}-2\cos{x}=0$, by assuming that $sin{x}\neq{0}$, we get the solution $x=\tan^{-1}(1/7)$. However, for the equation $14\sin{x}-2\cos{x}=0$, we have not yet rule out the possibility that $sin{x}=0$ since we previously assume that $sin{x}\neq{0}$. How can we show that x for which $\sin{x}=0$ is not a solution of $14\sin{x}-2\cos{x}=0$ and hence not a solution of the original equation?
    Why do you care if $\sin x=0$ at all? You should be caring about $\cos x \neq 0$:

    $7\sin x = \cos x$
    $7\dfrac{\sin x}{\cos x} = 1$
    $\tan x = \dfrac{1}{7}$

    You are dividing by $\cos x$. Fortunately, if $\cos x=0$, you already have a solution, so you can assume that $\cos x \neq 0$ and then solve.
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