Solve $\cos x-\sin(2x)=0$
I did:
$$\cos x=\color{blue}{\sin(\pi /2-x)}$$
therefore:
$$\color{blue}{\sin(\pi /2-x)}=\sin(2x)$$
$\endgroup$ 2Can I do that:??
now to solve only for $\pi/2-x=2x$
so $x=\pi/6+2\pi k$
5 Answers
$\begingroup$The first step is ok, but for the second we have that $$\sin A=\sin B\iff A+B=(2k+1)\pi\text{ or }A-B=2k\pi$$
Then, from the equality $\sin(\pi/2-x)=\sin 2x$ we get two sets of solutions:
- $\pi/2-x+2x=(2k+1)\pi$
- $2x-(\pi/2-x)=2k\pi$
Also, $$\cos x =\sin 2x \Rightarrow \cos x(1-2\sin x)=0 $$ $$ \cos x=0 \mid \sin x=\frac{1}{2}$$ where $\mid$ stands for "or". This can be easily solved.
Also, you can see that $\frac{5\pi}{6}$ is clearly a solution that is not included by your approach. Hope it helps.
EDIT: Expanding on @ajotatxe's one-liner on why $\sin x= \sin y \nRightarrow x=y$:
If $$\sin x-\sin y=0 \Rightarrow 2\cos \frac{x+y}{2}\sin \frac{x-y}{2} =0$$ Then solving, we have $$\frac{x+y}{2} = (2n+1)\frac{\pi}{2} \space\space\space\text{or}\space\space \frac{x-y}{2} = n\pi \space\space\space\text{where}\space\space n\in \mathbb Z$$ $$\Rightarrow x=(2n+1)\pi-y \space\space\space \text{or} \space\space x= 2n\pi + y \space\space\space\text{where}\space\space n\in \mathbb Z$$
$\endgroup$ $\begingroup$Slightly differently,
$$\cos x=\cos\left(\frac\pi2-2x\right)$$
yields
$$\pm x=\frac\pi2-2x+2k\pi$$ or
$$x=\frac{4k+1}{4\pm2}\pi.$$
$\endgroup$ $\begingroup$Cos x -sin(2x)= 0 implies that cos x- 2 sin x cos x=0 cos x (1- 2sin x) = 0
It means cos x=0 or sin x=1/2
Which gives answer x = 2n pi +- 90` or x= n pi +(-1)^n pi/6 I hope it helps you
$$\cos\left(x\right)-\sin\left(2x\right)=0\space\Longleftrightarrow\space\cos\left(x\right)=\cos\left(2x-\frac{\pi}{2}\right)$$
Take the inverse cosine of both sides:
$$x=2\pi\text{k}_1+2x-\frac{\pi}{2}\space\space\space\vee\space\space\space x=2\pi\text{k}_2-2x+\frac{\pi}{2}$$
Where $\text{k}_1\space\wedge\space\text{k}_2\in\mathbb{Z}$
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