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Q: How do you solve 2Sinx 1 equals 0?

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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.

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

4 sin2x = 1. Then, (2sinx)2 = 1, 2sinx = ±1, and sinx = ±½. Whence, x = 90° or 270°; or, in radians, x = π/2 or 3π/2.

3*sin2x = 2*sinx +1

9x2-9x = 0 x2-x = 0 x(x-1) = 0 x = 1 or x = 0

X = 0 (zero) Y = -1 (neg 1)

16x³ - x = 0 By factorization, we obtain: x(16x² - 1) = 0 x(4x - 1)(4x + 1) = 0 Set each term by 0 and solve for x to get: x = 0 and 4x - 1 = 0 and 4x + 1 = 0 x = {0, ±¼}

-1 plus 1 = 0 plus 1 =1

x2+1/4x = 0 x(x+1/4) = 0 Therefore: x = 0 or x = -1/4

2x2-x-3 = 0= 4 - x - 3 = 0= 1 - x = 0= 1 = xTherefore x = 1

cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...

x2-2x-3 = 0 (x+1)(x-3) = 0 x = -1 or x = 3

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