0
Sin(2x) + Sin(x) = 0
2Sin(x)Cos(x) + Sin(x) = 0
Factor 'Sin*(x)'
Sin(x) (2Cos(x) + 1 = 0
Hence
Sin(x) = 0
x = 0 , 180(pi) , 360(2pi) et.seq.,
&
2Cos(x) = -1
Cos(x) = -1/2 = -0.5
x = 120, 150, 480, et.seq.,
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Stevesin(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?
(1 + tanx)/sinxMultiply by sinx/sinxsinx + tanxsinxDivide by sin2x (1/sin2x) = cscxcscx + tan(x)csc(x)tanx = sinx/cosx and cscx = 1/sinxcscx + (sinx/cosx)(1/sinx)sinx cancels outcscx + 1/cosx1/cosx = secxcscx + secx
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.
