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What are the solutions to the trigonometric equation on the interval [0,2π)? 2cos^2x=√3 cosx

 Apr 9, 2020
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2cos^2  (x)  =  √3 cos x     divide both sides  by 2

 

cos^2 (x)  = (√3/2) cos x

 

cos^2 (x) - (√3/2) cos (x)  =   0       factor as

 

cos (x)  ( cos (x)  - (√3/2) )     =  0

 

Set  each factor to  0  and solve

 

cos x   =  0    and this happens  at x   = 0

 

And

 

cos  x - √3/2   =  0

 

cos x  = √3/2

 

And  this happens  at  x =  pi/6    and  11pi /6

 

 

So  the solutions  are  x  = 0 , x  = pi/6  and x  = 11pi/6

 

 

cool cool cool

 Apr 9, 2020

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