I am trying to solve this equation , but i get different answers from the book.
Can someone help me please?
How to solve
$\cos(x)-\cos(2x)+\cos(3x)=0$
My answers is : $x=45^\circ+90^\circ k$,
$x=\pm\frac{1}{2} +360^\circ k$
Book's answer: $\pm60^\circ+360^\circ k$ , $45^\circ+90^\circ k$
Best Answer
Notice, $$\cos x-\cos 2x+\cos 3x=0$$ $$(\cos x+\cos 3x)-\cos 2x=0$$ $$2\cos\left(\frac{x+3x}{2}\right)\cos\left(\frac{x-3x}{2}\right)-\cos 2x=0$$
$$2\cos 2x\cos x-\cos 2x=0$$ $$\cos 2x(2\cos x-1)=0 $$ Now, solving for $x$ as follows $$\cos 2x=0\implies 2x=(2k+1)\frac{\pi}{2}$$$$x=(2k+1)\frac{\pi}{4}$$$$ \color{red}{x=90^\circ k+45^\circ}$$ or $$2\cos x-1=0$$$$ \cos x=\frac{1}{2}=\cos \frac{\pi}{3}$$ $$x=2k\pi\pm\frac{\pi}{3}$$$$\color{red}{x=360^\circ k\pm 60^\circ}$$ Where, $\color{blue}{k}$ is any integer