0
x = 3pi/4
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What is the solution sets for sin3x plus sin2x plus sinx equals 0?
The solitions are in degrees. You may convert them to degrees should you wish. x= 0,90,120,180,240,270,360
What is the integral of cosx divided by sinx plus cosx from 0 to 2pi?
The Integral diverges. It has singularities whenever sin(x)+cos(x)=0. Singularities do not necessarily imply that the integral goes to infinity, but that is the case here, since the indefinite integral is x/2 + 1/2 Log[-Cos[x] - Sin[x]]. Obviously this diverges when evaluated at zero and 2pi.
What are the solutions of 2 cos squared x minus cos x equals 1?
2cos2x - cosx -1 = 0 Factor: (2cosx + 1)(cosx - 1) = 0 cosx = {-.5, 1} x = {...0, 120, 240, 360,...} degrees
How do you solve sin x - 2 sin x is equal to 0?
sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...
What is the derivative of the square root of 1-sinx?
√(1-sinx)=(1-sinx)1/2Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(1-sinx)1/2=(1/2)(1-sinx)1/2-1*d/dx(1-sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*d/dx(1-sinx)-The derivative of 1-sinx is:d/dx(u-v)=du/dx-dv/dxd/dx(1-sinx)=d/dx(1)-d/dx(sinx)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[d/dx(1)-d/dx(sinx)]-The derivative of 1 is 0 because it is a constant.-The derivative of sinx is:d/dx(sinu)=cos(u)*d/dx(u)d/dx(sinx)=cos(x)*d/dx(x)d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*d/dx(x))]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x)*1)]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[0-(cos(x))]d/dx(1-sinx)1/2=(1/2)(1-sinx)-1/2*[-cos(x)]d/dx(1-sinx)1/2=(-cosx)/[2√(1-sinx)]
Cos x plus sin x equals 0?
cosx + sinx = 0 when sinx = -cosx. By dividing both sides by cosx you get: sinx/cosx = -1 tanx = -1 The values where tanx = -1 are 3pi/4, 7pi/4, etc. Those are equivalent to 135 degrees, 315 degrees, etc.
How do you prove the following equation the quantity of sin theta divided by 1 minus cos theta minus the quantity 1 plus cos theta divided by sin theta equals 0?
You will have to bear with the angle being represented by x because this browser will not allow characters from other alphabets!sin^2x + cos^2x = 1=> sin^2x = 1 - cos^x = (1 + cosx)(1 - cosx)Divide both sides by sinx (assuming that sinx is not zero).=> sinx = (1 + cosx)(1 - cosx)/sinxDivide both sides by (1 - cosx)=> sinx/(1 - cosx) = (1 + cosx)/sinx=> sinx/(1 - cosx) - (1 + cosx)/sinx = 0
How would you find x when 0 equals 2sinxcosx-cosx?
2sinxcosx-cosx=0 Factored : cosx(2sinx-1)=0 2 solutions: cosx=0 or sinx=.5 For cosx=0, x=90 or 270 degrees For sinx=.5, x=30 degrees x = {30, 90, 270}
How does secx plus 1 divided by cotx equal 1 plus sinx divided by cosx?
secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = 1 + sinx/cosx, andsin/cos = tanx, therefore1/cosx + tanx = 1 + tanx, therefore1/cosx = 1, therfore1 = cosx.So, therfore, it is not neccesarily true.But if you meansecx plus 1 divided by cotx equals (1 plus sinx) divided by cosx(this is probably what you mean) Let's start over!secx = 1/cosxand 1/cotx = tanx, therefore1/cosx + tanx = (1+sinx)/cosx therefore1/cosx + tanx = 1/cosx + sinx/cosxsinx/cosx = tanx therfore1/cosx + tanx = 1/cosx + tanxDo you think this is correct? Subtract both sides by 1/cosx + tanx:0 = 0So, therefore, this is correct!(BTW, I'm in Grade 6! :P)
sinx+sin2x=0?
Sinx = 0 CosX= -1/2
What is the derivative of 1 divided by sinx?
y=1/sinxy'=(sinx*d/dx(1)-1*d/dx(sinx))/(sin2x)y'=(sinx*0-1(cosx))/(sin2x)y'=(-cosx)/(sin2x)y'=-(cosx/sinx)*(1/sinx)y'=-cotx*cscx
How do you break 1 sinx divided 1-cosx?
0
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
Is there any way to solve a system of equations with C and D as constants and x and y as variables sinx plus cozy - C equals 0 cosx plus siny - D equals 0?
0
What is a numerical value for x if cos2x plus 2sinx-2 equals 0?
cos2x + 2sinx - 2 = 0 (1-2sin2x)+2sinx-2=0 -(2sin2x-2sinx+1)=0 -2sinx(sinx+1)=0 -2sinx=0 , sinx+1=0 sinx=0 , sinx=1 x= 0(pi) , pi/2 , pi
What is the derivative of y equals -3xsinx - 1.5x to the 2 plus 5x when x equals pi?
y=-3x*sinx-1.5x2+5x, when x=πy'=d/dx(-3x*sinx)-d/dx(1.5x2)+d/dx(5x)y'=(-3x*d/dx(sinx)+sinx*d/dx(-3x))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx+sinx(-3))-d/dx(1.5x2)+d/dx(5x)y'=(-3x*cosx-3sinx)-3x+5y'=-3x*cosx-3sinx-3x+5 is the derivative at any point of that equation, now you only have to plug in π for xy'(π)=-3π*cosπ-3sinπ-3π+5y'(π)=-3π*(-1)-3(0)-3π+5y'(π)=3π-3π+5y'(π)=5
How do you find the solutions of tanx equals 2cscx?
tanx=2cscx sinx/cosx=2/sinx sin2x/cosx=2 sin2x=2cosx 1-cos2x=2cosx 0=cos2x+2cosx-1 Quadratic formula: cosx=(-2±√(2^2+4))/2 cosx=(-2±√8)/2 cosx=(-2±2√2)/2 cosx=-1±√2 cosx=approximately -2.41 or approximately 0.41. Since the range of the cosine function is [-1,1], only approx. 0.41 works. So: cosx= approx. 0.41 Need calculator now (I went as far as I could without one!) x=approx 1.148
