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The answer depends on x,
on whether x is measured in degrees or radians
and on whether the question is about sin(x) + 2 or sin(x + 2).
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How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
Differentiation of sin x 1 plus cosx?
The differentiation of sin x plus cosx is cos (x)-sin(x).
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
How do you solve 2 sin2-5 sin plus 2 equals 0?
2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.
What is the derivative of sin x plus 2?
d/dx [sin(x) + 2] = cos(x)
Sin x plus cos x?
sin(x) + cos(x) = sqrt(2) · sin(45°+x)
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
Differentiation of sin x 1 plus cosx?
The differentiation of sin x plus cosx is cos (x)-sin(x).
How do you differentiate sin squared x plus cos squared x?
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
Sine cubed plus cosine cubed divided by sine plus cosine?
[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).So now you just have:= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~Not doing any computation in this step.= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.= 1 - sin(x)cos(x)
How do you solve 2 sin2-5 sin plus 2 equals 0?
2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.
Sin x plus cos x equals?
The result is variant, therefore uncertain. The only sure thing is that sin(x)2 + cos(x)2 = 1.
