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

1. ## 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 $sin{x} = 0$?

2. ## 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 $0 \le x < \pi$ ...

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

$x = \arctan\left(\dfrac{1}{7}\right)$

3. ## Re: Solve 7 sin 2x - cos 2x = 0

Originally Posted by skeeter
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?

4. ## Re: Solve 7 sin 2x - cos 2x = 0

Originally Posted by woo
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 ...

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

5. ## Re: Solve 7 sin 2x - cos 2x = 0

Originally Posted by woo
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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