Could use help w/ solving 2 sin 4x = 2 cos 2x on [0, 2pi]

[pimarus]

I am working on a trig problem and I am stuck. Here is the problem:

2 sin 4x = 2 cos 2x where I am trying to find the solutions in the interval [0,2pi)

This is what I have so far:

I divided both sides by 2 and got: sin 4x = cos 2x
Since cos 2x = cos^2 (x) - sin^2 (x) , I now have sin 4x = cos^2 (x) - sin^2 (x)
Now since cos^2(x) = 1 - sin^2(x), I put that in and got:

sin 4x = 1 - sin^2(x) - sin^2(x)
sin 4x = 1 - 2sin^2(x)

This is where I get stuck. I don't know how to work with the sin 4x. If I change it to 2sin2xcos2x, that really doesn't help as now I have the cos back into the equation when I just trying to isolate for one function. Can anyone help me straighten this out?

Thank you!

 

[soroban]

Re: Could use help with a trig problem

Hello, pimarus

Solve: .$\displaystyle 2\sin 4x \:= \:2\cos 2x,\;\;0 \leq x \leq 2\pi$

I'd solve it this way . . .

$\displaystyle \text{Divide by 2: }\;\sin4x \:=\:\cos2x$

$\displaystyle \text{Then: }\quad\overbrace{2\sin2x\cos2x} \:=\:\cos2x \quad\Rightarrow\quad 2\sin2x\cos2x - \cos2x \:=\:0$

$\displaystyle \text{Factor: }\;\cos2x(2\sin2x - 1) \:=\:0$


$\displaystyle \cos2x \:=\:0\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2},\:\frac{5\pi}{2},\:\frac{7\pi}{2} \quad\Rightarrow\quad \boxed{x \:=\:\frac{\pi}{4},\:\frac{3\pi}{4},\:\frac{5\pi}{4},\:\frac{7\pi}{4}}$

$\displaystyle 2\sin2x - 1\:=\:0 \quad\Rightarrow\quad \sin2x \:=\:\frac{1}{2}\quad\Rightarrow\quad 2x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{13\pi}{6},\:\frac{17\pi}{6} \quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{12},\:\frac{5\pi}{12},\:\frac{13\pi}{12},\:\frac{17\pi}{12}}$

 

[pimarus]

Re: Could use help with a trig problem

That was awesome! It certainly made a lot more sense than the way I was heading. Thank you so much for taking the time to help me on this!