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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

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