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The Question was:

Express $2\cos{X} = \sin{X}$ in terms of $\sin{X}$ only.

I have had dealings with similar problems but for some reason, due to I believe a minor oversight, I am terribly vexed.

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    $\begingroup$ $\cos{x}^2 + \sin{x}^2 =1$ $\endgroup$ – Double AA May 23 '13 at 1:32
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    $\begingroup$ The key to these is to remember your trigonometric identities. You should find $\cos(x)^2 + \sin(x)^2 = 1$ useful here after solving for $\cos(x)$. $\endgroup$ – Kris Williams May 23 '13 at 1:34
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Note: too long for a comment...

Squaring to force the use of $\cos^2x+\sin^2x=1$ results in an equation which is not equivalent to the original equation. This actually creates another countable set of solutions.

First note that, $\cos \theta=0$ does not occur when the equation is fulfilled, so: $$(E): 2\cos x=\sin x \iff \tan x=2 $$ has the solution set $\arctan 2+\pi\mathbb{Z}$.

Now if you square $(E)$: $$ (E)^2: 4\cos^2x=\sin^2x\iff (2\cos x-\sin x)(2\cos x+\sin x)=0\iff\tan x=\pm2 $$ has the solution set $(\arctan 2 +\pi\mathbb{Z})\sqcup(-\arctan 2+\pi\mathbb{Z})$. So that's not equivalent to $(E)$.

Replacing $\cos x$ by $\pm\sqrt{1-\sin^2x}$ does not really work either. The best we can say is that $$ (E)\iff \cos x\geq 0 \wedge 2\sqrt{1-\sin^2 x}=\sin x\quad\mbox{or}\quad \cos x\leq 0 \wedge -2\sqrt{1-\sin^2 x}=\sin x. $$

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  • $\begingroup$ But $\sin(\tan^{-1} 2 + n\pi)=\sin(-\tan^{-1} 2 + (1-n)\pi)$, so the value in terms of $\sin(x)$ is the same either way. $\endgroup$ – Glen O May 23 '13 at 2:12
  • $\begingroup$ @GlenO ? I am just saying that the two equations are not equivalent. I am not talking about the values of $\sin x$ along the solutions. I am talking about $x$. $\endgroup$ – Julien May 23 '13 at 2:20
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Try using $\cos{x}^2 + \sin{x}^2 =1$ or some other known identity, but solving for $\cos{x}$ in terms of $\sin{x}$. Then just substitute.

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Given the equation $$2 \cos(x) = \sin(x)$$ and the instruction to write solely in terms of $\sin(x)$, I would begin by looking for an identity that involves $\cos(x)$, the term we want to transform and $\sin(x)$ the term we want to write everything in. This leaves us with the identity $$\cos(x)^2 + \sin(x)^2 =1.$$ We may then subtract $\sin(x)^2$ from both sides to have have $$\cos(x)^2 = 1 - \sin(x)^2.$$ As $\cos(x)^2$ is non-negative, we may take the square root of both sides and are left with $$\cos(x) = \pm \sqrt{1 - \sin(x)^2}.$$ We then replace the $\cos(x)$ in our equation with $\pm\sqrt{1 - \sin(x)^2}$ and thus we have two equations $$2 \sqrt{1 - \sin(x)^2} = \sin(x)$$ and $$-2 \sqrt{1 - \sin(x)^2} = \sin(x)$$

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    $\begingroup$ Of course, $\cos x$ could be negative... $\endgroup$ – Julien May 23 '13 at 1:46
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    $\begingroup$ -1 WHy did you choose the positive square root? $\endgroup$ – Calvin Lin May 23 '13 at 1:48
  • $\begingroup$ I have edited the answer to reflect using both roots. $\endgroup$ – Kris Williams May 23 '13 at 1:52
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As already mentioned by the users , you must use the identity :

$$\large \sin^2 x + \cos^2 x = 1 ........... \boxed{1} $$

Here is , how to start solving the problem :

$$\large \text{Given : } \quad 2\cos x = \sin x ......... \boxed{2}$$ Now, simplifying equation 1 further :

$$\large \cos^2 x = 1 - \sin^2 x $$ Solve for $\cos x $ further by taking square root both sides and put that value in equation 2 , you will get your answer.

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Of course, since $\sin^2(x)+\cos^2(x)=1$, if $2\cos(x)=\sin(x)$, squaring and adding $4\sin^2(x)$ to both sides yields $$ 4=5\sin^2(x)\tag{1} $$ Of course, $(1)$ also has solutions where $2\cos(x)=-\sin(x)$. Knowledge of $\sin(x)$ fully determines $|\cos(x)|$, but says nothing about the sign of $\cos(x)$. For this reason, the equation $2\cos(x)=\sin(x)$ cannot be fully characterized in terms of $\sin(x)$ alone.

On the other hand, $2\cos(x)=\sin(x)$ is equivalent to $\tan(x)=2$.

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