0
Add your answer:
Earn +20 pts
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
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.
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
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.
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.
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 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
Sinx plus cosx equals 0?
x = 3pi/4
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.
sinx+sin2x=0?
Sinx = 0 CosX= -1/2
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...
What is 1 sinx?
If you mean 1 - sinx = 0 then sinx = 1 (sin-1) x = 90
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.
When does cosx times sinx times sinx equal 1?
at the angles 0 and 360 degrees, or 0 and 2pi
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}
Prove this identity 1 plus cosx divide by sinx equals sinx divide by 1-cosx?
2
