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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.
Find all radian solutions in the interval 0 is less than or equal to x less than 2pi 2sin2 x - 6sinx - 1 equals 0?
2sin2x - 6sinx - 1 = 0Therefore, 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 radiansso x = 2pi - 0.1590 = 6.1242 radiansalso x = pi + 0.1590 = 3.3006 radians
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 is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
