Trig Question - Solve for x

[Scrutinize]

cos ( 2x ) = cos^2 ( (3x)/(2) ) Solve for x, without a specified interval.


I'm not really sure where to start, what I first did was I expanded the cos^2 ( (3x)/(2) ) into cos^2 ( (3x + 0)/(2) ) cos^2 ( (3x - 0)/(2) ) using the cos(a) + cos(b) identity I then simplified it into (cos(3x) + cos (0))/2 or (cos(3x) + 1)/2. So I had cos ( 2x) = (cos ( 3x ) + 1)/2

From there I just got lost on how to solve the question. Any help would be greatly appreciated, thanks!

 

[MarkFL]

I would continue and multiply by 2:

[MATH]2\cos(2x)=\cos(3x)+1[/MATH]
Now, using a double-angle and a triple-angle identity, I would write:

[MATH]2\left(2\cos^2(x)-1\right)=4\cos^3(x)-3\cos(x)+1[/MATH]
Or:

[MATH]4\cos^3(x)-4\cos^2(x)-3\cos(x)+3=0[/MATH]
Factor:

[MATH]4\cos^2(x)(\cos(x)-1)-3(\cos(x)-1)=0[/MATH]
[MATH]\left(4\cos^2(x)-3\right)(\cos(x)-1)=0[/MATH]
Can you proceed?

 

[MarkFL]

To follow up:

[MATH]\cos(x)=1\implies x=2\pi k[/MATH] where \(k\in\mathbb{Z}\)

[MATH]\cos(x)=\pm\frac{\sqrt{3}}{2}\implies x=\frac{\pi}{2}\pm\frac{\pi}{6}+\pi k=\frac{\pi}{6}(6k+3\pm1)[/MATH]

 

[Scrutinize]

I didn't know there was a triple-angle identity, that's what confused me the most I think. Thank you very much for the help!

 

[Scrutinize]

In hindsight, I could have done cos(2x + x) to simplify that and used the addition rule not knowing the triple angle identities existed.

 

[lev888]

In hindsight, I could have done cos(2x + x) to simplify that and used the addition rule not knowing the triple angle identities existed.

That's how the triple angle identity is derived.

 

[Scrutinize]

That's how the triple angle identity is derived.

Yea