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: .cos2A = 2cos2 A−1\displaystyle \cos2A \:=\:2\cos^2\!A - 1cos2A=2cos2A−1 I'll assume that: .0 ≤ θ ≤ 2π\displaystyle 0 \:\leq\: \theta\:\leq\:2\pi0≤θ≤2π cos2θ+cos4θ = 0\displaystyle \cos 2\theta + \cos 4\theta\:=\: 0cos2θ+cos4θ=0 Click to expand... Using the identity: .cos2θ+cos4θ⏟ = 0\displaystyle \cos2\theta + \underbrace{\cos4\theta} \;=\;0cos2θ+cos4θ=0 . . . . . . . . . . . cos2θ+2cos2 2θ−1⏞ = 0⇒2cos2 θ+cos2θ−1 = 0\displaystyle \cos2\theta + \overbrace{2\cos^2\!2\theta - 1} \:=\:0 \quad\Rightarrow\quad 2\cos^2\!\theta + \cos2\theta - 1 \:=\:0cos2θ+2cos22θ−1=0⇒2cos2θ+cos2θ−1=0 This factors: .(cos2θ+1)(2cos2θ−1) = 0\displaystyle (\cos2\theta + 1)(2\cos2\theta - 1) \;=\;0(cos2θ+1)(2cos2θ−1)=0 And we have two equations to solve: cos2θ+1 = 0⇒cos2θ = −1⇒2θ = π, 3π⇒θ = π2, 3π2\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}}cos2θ+1=0⇒cos2θ=−1⇒2θ=π,3π⇒θ=2π,23π 2cos2θ−1 = 0⇒cos2θ = 12⇒2θ = π3, 5π3, 7π3, 11π3⇒θ = π6, 5π6, 7π6, 11π6\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}}2cos2θ−1=0⇒cos2θ=21⇒2θ=3π,35π,37π,311π⇒θ=6π,65π,67π,611π
Hello, cledesma! We're expected to know this identity: .cos2A = 2cos2 A−1\displaystyle \cos2A \:=\:2\cos^2\!A - 1cos2A=2cos2A−1 I'll assume that: .0 ≤ θ ≤ 2π\displaystyle 0 \:\leq\: \theta\:\leq\:2\pi0≤θ≤2π cos2θ+cos4θ = 0\displaystyle \cos 2\theta + \cos 4\theta\:=\: 0cos2θ+cos4θ=0 Click to expand... Using the identity: .cos2θ+cos4θ⏟ = 0\displaystyle \cos2\theta + \underbrace{\cos4\theta} \;=\;0cos2θ+cos4θ=0 . . . . . . . . . . . cos2θ+2cos2 2θ−1⏞ = 0⇒2cos2 θ+cos2θ−1 = 0\displaystyle \cos2\theta + \overbrace{2\cos^2\!2\theta - 1} \:=\:0 \quad\Rightarrow\quad 2\cos^2\!\theta + \cos2\theta - 1 \:=\:0cos2θ+2cos22θ−1=0⇒2cos2θ+cos2θ−1=0 This factors: .(cos2θ+1)(2cos2θ−1) = 0\displaystyle (\cos2\theta + 1)(2\cos2\theta - 1) \;=\;0(cos2θ+1)(2cos2θ−1)=0 And we have two equations to solve: cos2θ+1 = 0⇒cos2θ = −1⇒2θ = π, 3π⇒θ = π2, 3π2\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}}cos2θ+1=0⇒cos2θ=−1⇒2θ=π,3π⇒θ=2π,23π 2cos2θ−1 = 0⇒cos2θ = 12⇒2θ = π3, 5π3, 7π3, 11π3⇒θ = π6, 5π6, 7π6, 11π6\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}}2cos2θ−1=0⇒cos2θ=21⇒2θ=3π,35π,37π,311π⇒θ=6π,65π,67π,611π
