what is the value of x in for which 2sin(1/2x)-cos(2x)=1?  

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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If the first term is sin(x/2), then we'll use the half angle identity:

sin(x/2) = sqrt[(1-cos x)/2]

We'll use the double angle identity for the 2nd term:

cos 2x = 2(cos x)^2 - 1

2sqrt[(1-cos x)/2] - 2(cos x)^2 + 1 = 1

We'll eliminate 1 both sides:

2sqrt[(1-cos x)/2] - 2(cos x)^2 = 0

We'll divide by 2:

sqrt[(1-cos x)/2] - (cos x)^2 = 0

We'll move (cos x)^2 to the right:

sqrt[(1-cos x)/2] = (cos x)^2

We'll raise to square:

(1-cos x)/2 = (cos x)^4

1-cosx - 2(cos x)^4 = 0

We'll write 2(cos x)^4 = (cos x)^4 + (cos x)^4

(1 - (cos x)^4) - cos x(1 + (cos x)^3) = 0

(1-cos x)(1+cos x)((1 + cos x)^4) - - cos x(1+cos x)(1 + cos x + (cos x)^2) = 0

(1 + cos x)[(1 - cos x)(1 + (cos x)^4) - cos x(1 + cos x +  (cos x)^2)] = 0

1 + cos x = 0

cos x = -1

x = pi

The solution of the equation in the interval (0 , 2pi) is x = pi.

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