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Find all values of x in the interval of [ 0 , 360 ) that satifsy each equation. Round approximatr answers to the nearest tenth of a degree.

 

 

1. sin x cos x + 2 cos x = 3 sin x + 6

 

2. 8 cos^4 x - 10 cos^2 x + 3 = 0

 

 

Thank you.

 Apr 30, 2019
 #1
avatar+106536 
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1.  sin x cos x + 2 cos x = 3 sin x + 6

 

sinxcosx + 2cos x - 3sinx - 6  = 0

 

(cosx -3) (sin x + 2)  = 0

 

No solutions    cosx  - 3  = 0   implies that   cos x  = 3   [impossible]

And       sin x + 2  = 0  implies that   sin x  = - 2   [ also impossible ]

 

 

cool cool cool

 Apr 30, 2019
 #2
avatar+106536 
+1

2.       8 cos^4 x - 10 cos^2 x + 3 = 0

 

(4cos^2x - 3) ( 2cos^2x - 1)  = 0

 

4cos^2x - 3  = 0

4cos^2x = 3

cos^2x = 3/4

cosx = √3/2       and this happens at  x  =  30°   and x = 330°

cos x = - √3/2    and this happens at  x = 150°   and x  =  210°

 

2cos^2x - 1  = 0

2cos^2x  = 1

cos^2 x  = 1/2

cos x = 1/√2      and this happens at x = 45°  and x = 315°

cos x = -1  /√2    and this happens at x =135°  and x = 225°

 

cool cool cool

 Apr 30, 2019

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