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(c) sin(2x) − 4 sin(x) = 0 (d) 4cos2(θ)−cos(θ)−2=0

Guest Dec 3, 2015

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 #1
avatar+78553 
+10

(c) sin(2x) − 4 sin(x) = 0     simplify

 

2sinxcosx - 4sinx = 0      factor

 

2sinx [ cosx - 2] = 0        the second factor has no solution

 

So

 

2sinx = 0      divide both sides by  2

 

sinx =  0      and this occurs at  [0 + n*180] °   where n is an integer

 

 

 

 

(d) 4cos2(θ)−cos(θ)−2=0       simplify

 

4[2cos^2 (θ) - 1] - cos (θ)  - 2  = 0

 

8cos^2 (θ) - cos (θ)  - 6   = 0

 

Let x = cos(θ)

 

8x^2 - x - 6  = 0

 

Using the quadratic formula, the solutions to this are  x = [1 - sqrt(193)] /16    and x = [1 + sqrt(193)]/ 16

 

So

 

cos (θ)  =  [1 - sqrt(193)] /16       or    cos (θ ) = [1 +sqrt(193)] /16

 

Taking the cosine inverse to find both angles, we have

 

(θ)  =   about 143.685°     and (θ)  = about 21.444°  

 

There are also two more angles on (0, 360)  at about 216.3°   and about  338.6°

 

Here's the graph of this one......it has an odd periodicity.......https://www.desmos.com/calculator/pwgqsyrurd

 

 

cool cool cool

CPhill  Dec 3, 2015
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1+0 Answers

 #1
avatar+78553 
+10
Best Answer

(c) sin(2x) − 4 sin(x) = 0     simplify

 

2sinxcosx - 4sinx = 0      factor

 

2sinx [ cosx - 2] = 0        the second factor has no solution

 

So

 

2sinx = 0      divide both sides by  2

 

sinx =  0      and this occurs at  [0 + n*180] °   where n is an integer

 

 

 

 

(d) 4cos2(θ)−cos(θ)−2=0       simplify

 

4[2cos^2 (θ) - 1] - cos (θ)  - 2  = 0

 

8cos^2 (θ) - cos (θ)  - 6   = 0

 

Let x = cos(θ)

 

8x^2 - x - 6  = 0

 

Using the quadratic formula, the solutions to this are  x = [1 - sqrt(193)] /16    and x = [1 + sqrt(193)]/ 16

 

So

 

cos (θ)  =  [1 - sqrt(193)] /16       or    cos (θ ) = [1 +sqrt(193)] /16

 

Taking the cosine inverse to find both angles, we have

 

(θ)  =   about 143.685°     and (θ)  = about 21.444°  

 

There are also two more angles on (0, 360)  at about 216.3°   and about  338.6°

 

Here's the graph of this one......it has an odd periodicity.......https://www.desmos.com/calculator/pwgqsyrurd

 

 

cool cool cool

CPhill  Dec 3, 2015

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