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Find all radian solutions in the interval 0 is less than or equal to x less than 2pi 2sin2 x - 6sinx - 1 equals 0?
Jkbtennis
2sin2x - 6sinx - 1 = 0
Therefore, using the quadratic equation,
sinx = (3-sqrt(11)/2 = -0.1583 or sinx > 3.
The latter solution is not possible since |sin(x)| cannot exceed 1.
arcsin(-0.1583) = -0.1590 radians
so x = 2pi - 0.1590 = 6.1242 radians
also x = pi + 0.1590 = 3.3006 radians
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Find all degree solutions 2sin2 6x plus 3sin6x plus 1 equals 0?
2sin2(6x) + 3sin(6x) + 1 = 0 Solving the quadratic, sin(6x) = -1 or sin (6x) = -0.5 sin(6x) = -1 => 6x = 45+60n degrees for integer n sin(6x) = -0.5 => 6x = 35+60n or 55+60n degrees for integer n.
Limit x approaches 0 sinx divide x equals 1 evaluate limit x approaches 0 cos2x-1 divide x?
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How do you find the solution for 2sin2 plus cos2 equals 1?
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Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What is the work for 2 sin squared x equals 2 plus cos x?
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Find all degree solutions 2sin2 6x plus 3sin6x plus 1 equals 0?
2sin2(6x) + 3sin(6x) + 1 = 0 Solving the quadratic, sin(6x) = -1 or sin (6x) = -0.5 sin(6x) = -1 => 6x = 45+60n degrees for integer n sin(6x) = -0.5 => 6x = 35+60n or 55+60n degrees for integer n.
What is 2sin2 in radians?
1.8185948536513634
Limit x approaches 0 sinx divide x equals 1 evaluate limit x approaches 0 cos2x-1 divide x?
== == Cos2x - 1 = [1 - 2sin2 x] - 1 = - 2sin2 x; so [Cos2x - 1] / x = -2 [sinx] [sinx / x] As x approaches 0, sinx / x app 1 while 2 sinx app 0; hence the limit is 0.
How do you find the solution for 2sin2 plus cos2 equals 1?
Assume the expression is: 2sin²(x) + cos²(x) = 1 Use this identity to work out the problem: cos²(x) = 1 - sin²(x) Then: 2sin²(x) + 1 - sin²(x) = 1 sin²(x) + 1 = 1 sin²(x) = 0 sin(x) = 0 Therefore, the solutions are: x = πk radians where k belongs to the set of integers In degrees, we obtain: x = 180l° where l belongs to the set of integers
Find the second derivative of y equals sin sqaured x?
y = sin2(x) y' = 2sin(x)cos(x) y'' = 2 [ cos(x)cos(x) + sin(x)(-sin(x)) ] = 2 [ cos2(x) - sin2(x) ] = 2 [ 1 - sin2(x) - sin2(x) ] = 2 [ 1 - 2sin2(x) ]
What does Sin squared x - Cos squared x equal?
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What is sine theta plus 2 cosine squared theta?
I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.I shall use x instead of theta since I cannot be bothered to paste it at each step.sin(x) + 2*cos2(x) = sin(x) + 2*[1 - sin2(x)] = sin(x) + 2 - 2sin2(x) which cannot be simplified further.
What is the work for 2 sin squared x equals 2 plus cos x?
2sin2 x2(1 - cos2 x) = 2 + cos x2 - 2cos2 x = 2 + cos x- 2cos2 x - cos x = 0cos x(-cos x - 1) = 0cos x = 0 or -cos x - 1 = 0cos x = -1Since the cosine has a period of 2pi, in the interval 0 ≤ x ≤ 2piwhen cos x = 0, x = pi/2 and x = 3pi/2, andwhen cos x = -1, x = pi.All values of x that satisfy the given equation are:x = pi/2 + 2npi, x = pi +2npi, and x = 3pi/2+ 2npi where n is any integer.
What is the solution to sec plus tan equals cos over 1 plus sin?
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What is the second derivative of y equals 2sinx cosx?
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How do you prove tan x plus tan x sec 2x equals tan 2x?
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