Solve 6 sin ( 2α)− 1 sin ( α ) = 0 for all solutions 0 ≤ α < 2 π. ?
回答 (8)
2021-02-10 1:09 pm
6 sin(2α) - sin(α) = 0
6 [2 sin(α) cos(α)] - sin(α) = 0
12 sin(α) cos(α) - sin(α) = 0
sin(α) [12 cos(α) - 1] = 0
sin(α) = 0 or cos(α) = 1/12
α = 0, π or α = 0.473, 2π - 0.473
Hence, α = 0, 0.473, π, 5.81
6 [2 sin(α) cos(α)] - sin(α) = 0
12 sin(α) cos(α) - sin(α) = 0
sin(α) [12 cos(α) - 1] = 0
sin(α) = 0 or cos(α) = 1/12
α = 0, π or α = 0.473, 2π - 0.473
Hence, α = 0, 0.473, π, 5.81
2021-02-10 2:23 pm
Rule
sin(2 α )= 2sin α cos α
sin(2 α )= 2sin α cos α
2021-02-11 1:04 am
Let A=alpha
6sin(2A)-sin(A)=0, 0=<A<=2pi.
=>
12sin(A)cos(A)-sin(A)=0
=>
sin(A)[12cos(A)-1]=0
=>
sin(A)=0=>A=0, pi, 2pi.
12cos(A)-1=0=>cos(A)=1/12=>A=1.48736624, 4.795819067.
Thus, the solution set is approximated to
{0, 1.487, 3.142, 4.796, 6.283}
6sin(2A)-sin(A)=0, 0=<A<=2pi.
=>
12sin(A)cos(A)-sin(A)=0
=>
sin(A)[12cos(A)-1]=0
=>
sin(A)=0=>A=0, pi, 2pi.
12cos(A)-1=0=>cos(A)=1/12=>A=1.48736624, 4.795819067.
Thus, the solution set is approximated to
{0, 1.487, 3.142, 4.796, 6.283}
2021-02-10 12:16 pm
Your notation is ambiguous.
Did you mean 6sin(2α) − sin(α) = 0 ?
Did you mean 6sin(2α) − sin(α) = 0 ?
2021-02-12 12:01 am
uhhhh the answer is 2
2021-02-10 5:47 pm
Let me replace Alpha by x. So Alpha = x
To evaluate x within Range 0 ≤ x < 2 π.
6 sin (2x) - sin x = 0
=> 6 sin x cos x - sin x = 0
=> sin x ( 6 cos x - 1 ) = 0
=> sin x = 0 and cos x = 1/6
When sin x = 0 , x = 0, π Radians in the given Range.
When cos x = 1/6 , x = 1.40 Radian, 4.88 Radian
x ( Or Alpha ) = 0, π , 1.40 Radian, 4.88 Radian .......... Answer
To evaluate x within Range 0 ≤ x < 2 π.
6 sin (2x) - sin x = 0
=> 6 sin x cos x - sin x = 0
=> sin x ( 6 cos x - 1 ) = 0
=> sin x = 0 and cos x = 1/6
When sin x = 0 , x = 0, π Radians in the given Range.
When cos x = 1/6 , x = 1.40 Radian, 4.88 Radian
x ( Or Alpha ) = 0, π , 1.40 Radian, 4.88 Radian .......... Answer
2021-02-10 2:49 pm
6sin(2α) - sin(α) = 0
12sin(α)cos(α) - sinα = 0
sin(α)[12cos(α) - 1] = 0
0 ≤ α < 2π
sin(α) = 0 → α = 0, π
12cos(α) - 1 = 0 → α = arccos(1/2), 2π - arccos(1/12)
α = 1.48737, 4.79582
Ans:
α = 0, 1.48737, π, 4.79582
12sin(α)cos(α) - sinα = 0
sin(α)[12cos(α) - 1] = 0
0 ≤ α < 2π
sin(α) = 0 → α = 0, π
12cos(α) - 1 = 0 → α = arccos(1/2), 2π - arccos(1/12)
α = 1.48737, 4.79582
Ans:
α = 0, 1.48737, π, 4.79582
2021-02-10 12:31 pm
6 sin (2α) - 1 sin (α) = 0
6 (2 sin α cos α) - sin α - 0
sin α (12 cos α - 1) - 0
⇒ Either sin α = 0 or 12 cos α -1 = 0
case 1. sin α = 0 ⇒ α = 0, π, 2π ..................ANS
case 2. 12 cos α - 1 = 0
⇒ cos α = 1/12
α = cos⁻¹ (1/12)
α = 0.47π .......................ANS
6 (2 sin α cos α) - sin α - 0
sin α (12 cos α - 1) - 0
⇒ Either sin α = 0 or 12 cos α -1 = 0
case 1. sin α = 0 ⇒ α = 0, π, 2π ..................ANS
case 2. 12 cos α - 1 = 0
⇒ cos α = 1/12
α = cos⁻¹ (1/12)
α = 0.47π .......................ANS
收錄日期: 2021-04-24 08:13:17
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