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cos4θ +cos6θ =0

define θ

 Jan 27, 2017
 #1
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cos4θ +cos6θ =0

 

cos4θ  =  cos(5θ - θ)   =   cos5θcosθ + sin5θsitnθ

cos6θ =   cos(5θ + θ)  =   cos5θcosθ - sin5θsinθ

 

So.......

 

cos4θ +cos6θ = 0  →  [ cos5θcosθ + sin5θsinθ ] + [cos5θcosθ - sin5θsinθ] = 0 →

 

2cos5θcosθ  = 0      divide by 2

 

cos5θcosθ= 0

 

Set  each factor to  0  

cosθ  = 0   at θ  =    pi/2  + n*pi    ..... and........

cos5θ = 0   at  θ  =  pi/10 + n*pi/5

 

Here's a graph of the solutions :

 

 

cool cool cool

 Jan 27, 2017

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