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1) Solve the following equation, where theta is in the reals.

4cos(2theta)+3=0

 

 

2)Solve for x over the interval of [0, 2pi]

-2 tan^2x=0

 

 

 

Please include a little bit of explanation, thanks

 Dec 31, 2018
 #1
avatar+98005 
+2

4cos(2 theta)+3=0     subtract   3 from both sides

 

4 cos (2 theta)  =  - 3     divide both sides by 4

 

cos (2 theta)  = -3 / 4

 

cos (theta)  =  -3/4    at  ≈ [ 2.42   ±  2pi ]  rads  and at ≈ [3.86 ± 2pi ]  rads

 

So

 

cos ( 2 theta) =   (  [ 2.42 ± 2pi ] / 2   )  rads   and   at   ( [ 3.86 ± 2pi ] / 2 rads )  =

 

[ 1.21 ± pi ] rads    and   [ 1.93 ± pi ]  rads

 

 

cool cool cool

 Dec 31, 2018
edited by CPhill  Dec 31, 2018
 #2
avatar+98005 
+2

2)Solve for x over the interval of [0, 2pi]

-2 tan^2x=0

 

Divide both sides by - 2

 

tan^2  x  = 0        take the square root and we get that

 

tan x  =  0

 

This happens at   x =   0 ,  pi  and 2pi    rads     on the requested interval

 

 

cool cool cool

 Dec 31, 2018
 #3
avatar+98005 
+1

Comet's second queston is actually this :

 

2)Solve for x over the interval of [0, 2pi]

-2 + tan^2x=0

 

Add 2 to both sides

 

tan^2x  =  2       taske both square roots

 

tan x = ± √2

 

We either have that    

 

tan x = √2       

 

Take the tan inverse to find the angle in  rads 

 

arcctan (√2)  ≈ .955 rads    this will also occur at   pi +  .955 rads = 4.096 rads

 

Also

 

tan x = -√2

 

arctan ( -√2)  ≈ -.955 rads     =   [2pi -  .955] rads  ≈ 5.328 rads

 

And once more at  [5.328 - pi] rads   ≈ 2.19 rads   

 

So....the four answers are ≈   [ .955 , 2.19, 4.096 and 5.328 ]   rads

 

 

cool cool cool

 Dec 31, 2018

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