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x = 3pi/4

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โˆ™ 2010-02-06 18:26:55
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A polynomial of degree zero is a constant term

The grouping method of factoring can still be used when only some of the terms share a common factor A True B False

The sum or difference of p and q is the of the x-term in the trinomial

A number a power of a variable or a product of the two is a monomial while a polynomial is the of monomials

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Q: Sinx plus cosx equals 0
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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 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 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 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)

What is the derivative of 1 divided by sinx?


How do you break 1 sinx divided 1-cosx?


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?


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

When does Cosx equals 1?

when the angle is 0 degrees

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

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

Solve 2sinx-sin3x equals 0?

2sinx - sin3x = 0 2sinx - 3sinx + 4sin3x = 0 4sin3x - sinx = 0 sinx(4sin2x - 1) = 0 sinx*(2sinx - 1)(2sinx + 1) = 0 so sinx = 0 or sinx = -1/2 or sinx = 1/2 It is not possible to go any further since the domain for x is not defined.

What is the transformation that maps y equals sinx onto y equals the inverse of sinx?

f(x) = 1/x except where x = 0.

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

Is the solution of 1 plus cosx equals 0 is ฯ€ plus 2kฯ€?

Yes, that looks good. That's 180 degrees plus every multiple of 360 degrees more.

Sin2x - radical 2 cosx equals 0?

Sin2x = radical 2

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

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.

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 solve sin x - 2 sin x is equal to 0?

sinx(1-sinx)=0 sinx=0 or 1 x= 0, 90, 180, 270, 360...

How do you construct the Lagrange interpolation for a polynomial cosx plus sinx given the four values of x?

Construct the Lagrange interpolating polynomial P1(x) for f(x) = cos(x)+sin(x) when x0 = 0; x1 = 0:3. Find the absolute error on the interval [x0; x1].

What is 1- sinx?

If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90

Verify the identity sinx cotx - cosx divided by tanx equals 0?

(sin(x)cot(x) - cos(x))/tan(x)(Multiply by tan(x)/tan(x))sin(x) - cos(x)tan(x)(tan(x) = sin(x)/cos(x))sinx - cos(x)(sin(x)/cos(x))(cos(x) cancels out)sin(x) - sin(x)0

How do you solve sin squared theta plus cos theta equals sin theta plus cos squared theta?

For simplicity's sake, X represent theta. This is the original problem: sin2x+ cosX = cos2X + sinX This handy-dandy property is key for all you trig fanatics: sin2x+ cos2x = 1 With this basic property, you can figure out that sin2 x=1-cos2x and cos2x= 1-sin2x So we can change the original problem to: 1-cos2x+cosx = 1-sin2X + sinX -cos2x + cosx =-sin2x + sinX Basic logic tells you that one of two things are happening. sin2x is equal to sinx AND cos2x is equal to cosx. The only two numbers that are the same squared as they are to the first power are 1 and 0. X could equal 0, which has a cosine of 1 and a sine of 0, or it could equal pi/2, which has a cosine of 0 and a sine of 1. The other possibility whatever x (or theta) is, it's sine is equal to its cosine. This happens twice on the unit circle, once at pi/4 and once at 5pi/4. If you're solving for all possible values for x and not just a set range on the unit circle, then the final solution is: x=0+2pin x=pi/2+2pin x= pi/4 +2pin x=5pi/4+2pin (note that n is a variable)