I have to solve $$\cos(3x)= \cos(2x)$$
I found how to express both of them for $x$ only and now I have $$4\cos^3(x)-3\cos(x)=2\cos^2(x)-1$$
What do I do now?
Asked
Active
Viewed 5,504 times
11
-
1my hint: $\cos A=\cos B\iff A=B$ – Bhaskara-III Mar 09 '16 at 13:17
6 Answers
17
Notice, we have $$\cos(3x)=\cos(2x)$$ $$3x=2n\pi\pm 2x$$ Where, $n$ is any integer
Now, we have the following solutions
$$3x=2n\pi+2x\implies \color{red}{x=2n\pi}$$ or $$3x=2n\pi-2x\implies \color{red}{x=\frac{2n\pi}{5}}$$
Harish Chandra Rajpoot
- 37,450
-
I see that it works pretty well, thank you! But can you explain the +/- 2x? – Rosalie Landry-Ouellet Sep 25 '15 at 19:35
-
6Alright, we know that $\cos(2n\pi+\theta)=\cos(2n\pi-\theta)=\cos\theta$ Hence, the general solution of $\cos (3x)=\cos(2x)$ is $3x=2n\pi\pm 2x$ – Harish Chandra Rajpoot Sep 25 '15 at 19:41
-
7It appears that your first family of solutions is a subset of your second family. – paw88789 Sep 25 '15 at 19:43
6
We have the formula $$ \cos{(a+b)}-\cos{(a-b)} = -2\sin{a}\sin{b}, $$ so if we put $a=5x/2,b=x/2$, we have $$ 0 = \cos{3x}-\cos{2x} = -2\sin{\left(\frac{5}{2}x\right)}\sin{\left(\frac{x}{2}\right)},$$ so you just have to determine the $x$ for which either one of the sines is equal to $0$, which I'm sure you can do.
Chappers
- 67,606
2
You reached to the cubic equation $$4t^3-2t^2-3t+1=0$$ where $$t=\cos(x)\in [-1,1]$$
One root is $t=1$. Divide the polynomial by $(t-1)$ and find the other two roots. Finally plug in $t=\cos(x)$ to find $x$.
Svetoslav
- 5,205
- 2
- 16
- 35
0
hint: $$\cos \alpha=\cos \beta$$$$\iff \alpha=2k\pi\pm \beta$$ where; $k=0, \pm1, \pm2, \pm 3, \ldots$
Bhaskara-III
- 1,571
0
$$\cos(3x)=\cos(2x)$$ $$\cos(3x)-\cos(2x)=0$$ $$2\sin(3x/2)\sin(-x/2)=0$$ $$-2\sin(3x/2)\sin(x/2)=0$$ $$\sin(3x/2)=0\implies 3x/2=k\pi\iff x=2k\pi/3$$ or $$\sin(x/2)=0\implies x/2=k\pi\iff x=2k\pi$$
jeanne clement
- 670