0
sin(2x) + sin(x) = 0
2sin(x)cos(x) + sin(x) = 0
sin(x)[2cos(x) + 1] = 0
sin(x) = 0 OR 2cos(x) + 1 = 0
sin(x) = 0 OR cos(x) = -1/2
x = n*pi OR x = 2/3*pi + 2n*pi OR x = -2/3*pi + 2n*pi
x = pi*[2n + (0 OR 2/3 OR 1 OR 4/3)]
Note that n may be any integer.
The solutions in [-2pi, 2pi] are: -2pi, -4/3pi, -pi, -2/3pi, 0, 2/3pi, pi, 4/3pi, 2pi
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Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
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Can you Show 1 over sinx cosx - cosx over sinx equals tanx?
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
What is the square root of sin2x and why?
sin2x is the conventional way of writing (sinx)2; it does not denote the sine of sinx as one might expect. So the square root is just sinx.
Tan plus cot divided by tan equals csc squared?
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
Parenthesis 1 plus tanx end parenthesis divided by sinx equals cscx plus secx?
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How do you simplify cosx plus sinx tanx?
to simplify Cosx=Sinx Tanx you should remember your fundamental and pythagorean identities.. Cosx + Sinx Tanx Cosx + Sinx (Sinx/Cosx) <---------- From Tanx= Sinx/Cosx Cosx + Sin2x/ Cos x <------------- do the LCD Cosx (Cosx/Cosx) + Sin2x/Cosx (Cos2x+Sin2x)/Cosx 1/Cosx <--------- From Sin2x + Cos2x =1 or Secx <-------- answer Comment if you have questions...:))
How should I solve 3sin2x 2sinx 1?
3*sin2x = 2*sinx +1
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 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
Can you Show 1 over sinx cosx - cosx over sinx equals tanx?
From the Pythagorean identity, sin2x = 1-cos2x. LHS = 1/(sinx cosx) - cosx/sinx LHS = 1/(sinx cosx) - (cosx/sinx)(cosx/cosx) LHS = 1/(sinx cosx) - cos2x/(sinx cosx) LHS = (1- cos2x)/(sinx cosx) LHS = sin2x /(sinx cosx) [from Pythagorean identity] LHS = sin2x /(sinx cosx) LHS = sinx/cosx LHS = tanx [by definition] RHS = tanx LHS = RHS and so the identity is proven. Q.E.D.
How do you solve 1 minus cosx divided by sinx plus sinx divided by 1 minus cosx to get 2cscx?
(1-cosx)/sinx + sinx/(1- cosx) = [(1 - cosx)*(1 - cosx) + sinx*sinx]/[sinx*(1-cosx)] = [1 - 2cosx + cos2x + sin2x]/[sinx*(1-cosx)] = [2 - 2cosx]/[sinx*(1-cosx)] = [2*(1-cosx)]/[sinx*(1-cosx)] = 2/sinx = 2cosecx
What is the square root of sin2x and why?
sin2x is the conventional way of writing (sinx)2; it does not denote the sine of sinx as one might expect. So the square root is just sinx.
sinx+sin2x=0?
Sinx = 0 CosX= -1/2
Tan plus cot divided by tan equals csc squared?
(tanx+cotx)/tanx=(tanx/tanx) + (cotx/tanx) = 1 + (cosx/sinx)/(sinx/cosx)=1 + cos2x/sin2x = 1+cot2x= csc2x This is a pythagorean identity.
How do you solve 2 cos squared x - sinx - 1?
1
What is the differenciation of sinx plus sin2x?
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