Can you please help me solve this problem. cos(2theta) + cos(4theta)= 0
C cledesma New member Joined Mar 5, 2009 Messages 1 Mar 5, 2009 #1 Can you please help me solve this problem. cos(2theta) + cos(4theta)= 0
S soroban Elite Member Joined Jan 28, 2005 Messages 5,586 Mar 5, 2009 #2 Hello, cledesma! We're expected to know this identity: .\(\displaystyle \cos2A \:=\:2\cos^2\!A - 1\) I'll assume that: .\(\displaystyle 0 \:\leq\: \theta\:\leq\:2\pi\) \(\displaystyle \cos 2\theta + \cos 4\theta\:=\: 0\) Click to expand... Using the identity: .\(\displaystyle \cos2\theta + \underbrace{\cos4\theta} \;=\;0\) . . . . . . . . . . . \(\displaystyle \cos2\theta + \overbrace{2\cos^2\!2\theta - 1} \:=\:0 \quad\Rightarrow\quad 2\cos^2\!\theta + \cos2\theta - 1 \:=\:0\) This factors: .\(\displaystyle (\cos2\theta + 1)(2\cos2\theta - 1) \;=\;0\) And we have two equations to solve: \(\displaystyle \cos2\theta + 1 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:-1 \quad\Rightarrow\quad 2\theta \:=\:\pi,\:3\pi \quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}\) \(\displaystyle 2\cos2\theta - 1 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{3},\:\frac{5\pi}{3},\:\frac{7\pi}{3},\:\frac{11\pi}{3}\quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{7\pi}{6},\:\frac{11\pi}{6}}\)
Hello, cledesma! We're expected to know this identity: .\(\displaystyle \cos2A \:=\:2\cos^2\!A - 1\) I'll assume that: .\(\displaystyle 0 \:\leq\: \theta\:\leq\:2\pi\) \(\displaystyle \cos 2\theta + \cos 4\theta\:=\: 0\) Click to expand... Using the identity: .\(\displaystyle \cos2\theta + \underbrace{\cos4\theta} \;=\;0\) . . . . . . . . . . . \(\displaystyle \cos2\theta + \overbrace{2\cos^2\!2\theta - 1} \:=\:0 \quad\Rightarrow\quad 2\cos^2\!\theta + \cos2\theta - 1 \:=\:0\) This factors: .\(\displaystyle (\cos2\theta + 1)(2\cos2\theta - 1) \;=\;0\) And we have two equations to solve: \(\displaystyle \cos2\theta + 1 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:-1 \quad\Rightarrow\quad 2\theta \:=\:\pi,\:3\pi \quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}\) \(\displaystyle 2\cos2\theta - 1 \:=\:0 \quad\Rightarrow\quad \cos2\theta \:=\:\frac{1}{2} \quad\Rightarrow\quad 2\theta \:=\:\frac{\pi}{3},\:\frac{5\pi}{3},\:\frac{7\pi}{3},\:\frac{11\pi}{3}\quad\Rightarrow\quad \boxed{\theta \:=\:\frac{\pi}{6},\:\frac{5\pi}{6},\:\frac{7\pi}{6},\:\frac{11\pi}{6}}\)
