0
OK. Let's give it a whirl:
2 cos2 + sin - 1 = 0
Remember that cos2 = 1 - sin2 . Then, substitute that in the given equation
to see whether there appears to be any chance that it makes life easier:
2 (1 - sin2) + sin - 1 = 0
Eliminate parentheses :
2 - 2 sin2 + sin - 1 = 0
Combine like terms, clean it up, and multiply each side by -1 :
2 sin2 - sin - 1 = 0
This looks factorable:
(2 sin + 1) (sin - 1) = 0
Now we have it within our grasp.
2 sin(x) = 0
sin(x) = 0
x = + or - N pi
sin(x) - 1 = 0
sin(x) = 1
x = pi/2 + 2N pi
'x' is 90 degrees, and every multiple of 180 degrees.
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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 value of sinX wrt cosX?
sin(x) = sqrt[ 1 - cos2(x) ]
Sinx plus cosx equals 0?
x = 3pi/4
Solve 6sinx equals 1 plus 9sinx algebraically over the domain 0 is greater than or equal to x is less than 2pi?
6*sinx = 1 + 9*sinx => 3*sinx = -1 => sinx = -1/3Let f(x) = sinx + 1/3then the solution to sinx = -1/3 is the zero of f(x)f'(x) = cosxUsing Newton-Raphson, the solutions are x = 3.4814 and 5.9480It would have been simpler to solve it using trigonometry, but the question specified an algebraic solution.
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
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
How do you solve 6sin x 1 plus 9sin x algebraically over the domain 0 x 2pi?
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Verify that Cos theta cot theta plus sin theta equals csc theta?
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f(x) = 1/x except where x = 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 reverse of sin2x - sinx-1 equals 0?
There is not a "reverse" - whatever that may mean. The solution is x = (-0.6662 + 2k*pi) radians where k is an integer.
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
