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Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3
Let f(x) = sinx + 1/3
then the solution to sinx = -1/3 is the zero of f(x)
f'(x) = cosx
Using Newton-Raphson, the solutions are x = 3.4814 and 5.9480
It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
What else can I help you with?
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
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
What is the amplitude for the function y equals 6sinx?
The amplitude of the function [ sin(x) ] is 1 peak and 2 peak-to-peak . The amplitude of 6 times that function is 6 peak and 12 peak-to-peak.
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
Integral of (sin2x)(cos3x)?
Int[sin(2x)*cos(3x)]dx = Int[(2sinx*cosx)*(4cos^3x - 3cosx)]dx= Int[(8sinx*cos^4x - 6sinx*cos^2x)]dx Let cosx = u then du/dx = -sinx So, the integral is Int[-8*u^4 + 6*u^2]du = -8/5*u^5 + 2u^3 + c where c is a constant of integration = -8/5*cos^5x + 2cos^3x + c
