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Solve sin 2 x = 0.8193, when 0≤ x ≤ 360˚?

2006-12-18 15:43:25 · 9 answers · asked by Anonymous in Science & Mathematics Mathematics

9 answers

2x = arcsin(0.8193) + 2n*pi rad = 0.96 + 2n*pi rad or -0.96+pi +2n*pi rad

x = (0.48 + n*pi) rad or (-1.09 + n*pi) rad
or, in degrees,

x = 55.014 + 180 n or x = -124.985 + 180 n
n is any integer.
Now, choose the n values that make your angle reside in the given interval [0,360].

Take n = 0, to get
x = 55.014 or x = -124.985

2006-12-18 15:51:27 · answer #1 · answered by mulla sadra 3 · 0 0

The line y = 0.8193 intersects the graph of y = sin x in 2 places in the interval [0, 360], once at x = 55.01° and again at x = 124.99°, which is 90° ± 34.99°.

But the line intersects y = sin(2x) in 4 places, 45° ± 17.49° AND 225° ± 17.49°.

The general solution to sin x = .8193 is a pair of lists, x = 55.01° + 360n and x = (180° - 55.01°) + 360n. So when you divide 2x = each list by 2, you get

x = 27.5° + 180n = 27.5, 207.5, .... and
x = 90 - 27.5 + 180n = 62.5, 242.5, ....

and the 1st 2 in each list are between 0 and 360.

2006-12-18 16:25:12 · answer #2 · answered by Philo 7 · 0 0

Sin 2x = 0.8193 = Sin (55 deg)
If Sin A = Sin B,
the generalized values of A is >> A = n times (pie) + (-1)^n times (B)
pie = 180 deg
n = 0, +/-1, +/-2, +/-3, .......

We have, Sin 2x = Sin 55
2x = n times (pie) + (-1)^n times 55
x = n times 180/2 + (-1)^n times 55/2

x = n times 90 + (-1)^n times 27 1/2 deg

Given: 0 ≤ x ≤ 360˚

Plugging n = 0, 1, 2, 3, 4 ...
x = 27.5˚, 90-27.5 = 62.5˚, 180+27.5 = 207.5˚, 270-27.5 = 242.5˚

2006-12-18 16:17:30 · answer #3 · answered by Sheen 4 · 0 0

Sin²(x) + 2Sin(x)Cos(x) = 0 ingredient Sin(x)(Sin(x) + 2Cos(x)) = 0 So the two Sin(x) = 0 or Sin(x) + 2Cos(x) = 0 Sin(x) = 0 has ideas x = 0 + n?, the place n is an integer Sin(x) + 2Cos(x) = 0 ? Sin(x) = -2Cos(x) or Sin(x)/Cos(x) = Tan(x) = -2 So x = ArcTan(-2) = -a million.107… + n? the place n is an integer. verify this bring about the unique equation(s), I did!

2016-10-15 05:25:57 · answer #4 · answered by ? 4 · 0 0

sin2x=.8193
2x=arcsin.8193
basic angle=55.01
alter range:
0 ≤ 2x ≤ 720
sine +ve in 1st,2nd,5th,6th quadrants:
2x=55.01, 180-55.01, 360+55.01, 540-55.01
therefore
x= 27.51, 62.50, 207.51, 242.50

2006-12-18 18:11:19 · answer #5 · answered by Maths Rocks 4 · 0 0

sin 2 x = 0.8193
2x=sin(_1)(0.8193)
2x=55.01478
x=(55.01478)/2
x=27.50739

also x=27.50739+180=207.50739

2006-12-18 15:54:58 · answer #6 · answered by angel 2 · 0 0

2x=arcsin(0.8193)
x=1/2(arcsin0.8193)
or pi-[arcsin0.8193]/2

2006-12-18 15:58:52 · answer #7 · answered by raj 7 · 0 0

2x = arcsin(0.8193) = 55.015° or 124.99°
x = 27.507°, 62.493°, 207.51° or 291.5°

2006-12-18 16:03:18 · answer #8 · answered by Helmut 7 · 0 0

x=27.5 degrees or 62.49

2006-12-18 15:48:45 · answer #9 · answered by Anonymous · 0 0

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