## satellite73 4 years ago i have a question. why is the solution to $sin(3x)=\sin(2x)$ $x=\frac{\pi}{5}, x=\frac{3\pi}{5}$?

1. anonymous

i mean i worked it out and found this answer, but the last step of my solution required finding $\cos^{-1}\left (\frac{1-\sqrt{5}}{4}\right)$ so i am wondering if there is a snappy way to do it, because the answer is so neat and clean.

2. anonymous

3. phi

@sat my instinct was to take your approach. But looking further, I see sin(a) - sin(b)= 2 cos(0.5(a+b)) sin(0.5(a-b)) gets us to the answer in a snappy way

4. anonymous

@phi, wow it sure does, doesn't it!

5. anonymous

for $\cos x = \frac{1+ \sqrt5}{4}$ and $\cos x = \frac{1- \sqrt5}{4}$ Since x=pi/5 is a solution to $\cos x = \frac{1+ \sqrt5}{4}$ x=pi/5+2kpi are also solutions (k is an integer) also, x=-pi/5+2kpi are also solutions. since cos(-x)=cos(x) Similarly since x=3pi/5 is a solution to $\cos x = \frac{1- \sqrt5}{4}$ x=3pi/5+2kpi are also solutions (k is an integer) also, x=-3pi/5+2kpi are also solutions. since cos(-x)=cos(x) And for sin (x) = 0 we have x = 0 + kpi are solutions. Therefore, the solutions to the equation sin(2x)=sin(3x) are $x = k \pi$ $x=2k \pi \pm \frac{\pi}{5}$ and $x=2k \pi \pm \frac{3\pi}{5}$

6. anonymous

I'm only continuing on satellite73's response in the attached link.

7. phi

@tomas: replace sin^2(x) with 1-cos^2 x you get sat's equation

8. anonymous

|dw:1327764356757:dw|

9. anonymous

it means 2n$2n \Pi + -1^{n} \times b$

10. anonymous

phi's answer is short and sweet and makes it clear, but i am still wondering why, when you take $\cos^{-1}\left(\frac{1-\sqrt{5}}{4}\right)=\frac{\pi}{5}$

11. anonymous

i mean other than the fact that it is true. why is $\cos(\frac{\pi}{5})$ the same as the solution to $4x^2-2x-1=0$