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

Guest Dec 3, 2015

#1
+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

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

CPhill  Dec 3, 2015
Sort:

#1
+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

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

CPhill  Dec 3, 2015

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