how many roots

Geometry Level pending

cos ⁡ ( x ) + cos ⁡ ( 2 x ) + cos ⁡ ( 5 x ) = 0 \cos (x) + \cos (2x) + \cos (5x) = 0

Find the number of roots for x ∈ [ 0 , π ] x \in [0, \pi] .


The answer is 5.

Aly Ahmed by Aly Ahmed , 34, Egypt

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1 solution

cos x + cos ( 2 x ) + cos ( 5 x ) = 0 ⇒ cos x + cos ( 5 x ) + cos ( 2 x ) = 0 ⇒ 2 cos ( 3 x ) cos ( 2 x ) + cos ( 2 x ) = 0 ⇒ cos ( 2 x ) [ 2 cos ( 3 x ) + 1 ] = 0 ⇒ cos ( 2 x ) = 0 \hspace{12pt}\text{cos }x + \text{cos }(2x) + \text{cos}(5x) = 0\newline\Rightarrow \text{cos }x + \text{cos }(5x) + \text{cos }(2x) = 0\newline\Rightarrow 2\text{cos}(3x)\text{ cos}(2x) + \text{cos}(2x) = 0 \newline\Rightarrow \text{cos}(2x)[2\text{cos}(3x) + 1] = 0\newline\Rightarrow \text{cos}(2x) = 0\hspace{30pt} or 2 cos ( 3 x ) + 1 = 0 \hspace{30pt}2\text{cos}(3x) + 1 = 0

2 x = ( 2 n − 1 ) π 2 2x = \Large\frac{(2n-1)\pi}{2}\hspace{30pt} or 3 x = 2 n π ± 2 π 3 \hspace{30pt}3x = 2n\pi \pm\Large\frac{2\pi}{3}

x = π 4 x = \large\frac{\pi}{4} , 3 π 4 ,\large\frac{3\pi}{4}\hspace{30pt} or x = 2 π 9 \hspace{30pt} x = \large\frac{2\pi}{9} , 4 π 9 , \large\frac{4\pi}{9} , 8 π 9 ,\large\frac{8\pi}{9}

So there are 5 5 solutions.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

I don't understand the third line, please help! How does 2 cos ⁡ ( 3 x ) cos ⁡ ( 2 x ) = cos ⁡ ( x ) + cos ⁡ ( 5 x ) 2\cos{(3x)} \cos{(2x)} = \cos{(x)} + \cos{(5x)}

Mahdi Raza - 1 year ago

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Using the formula,

cos ( C ) + cos ( D ) = 2 cos ( C + D 2 ) \text{cos}(C) + \text{cos}(D) = 2\text{cos}\Large\big(\frac{C + D}{2}\big) cos ( C − D 2 ) \text{cos}\Large\big(\frac{C-D}{2}\big)

Shikhar Srivastava - 1 year ago

Thank you @Pi Han Goh and @Shikhar Srivastava , I will check it out. Looks quite interesting! Thanks!

Mahdi Raza - 1 year ago

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