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How do you find all solutions of

+1 vote

 

sin2x = sin^(1/2)x

x in the interval [0, 2π]?

 

asked Feb 14, 2013 in TRIGONOMETRY by payton Apprentice

1 Answer

+1 vote

sin(2x) = √[sin(x)]

Take out square each side.

{sin(2x)}2 = {√[sin(x)]}2

Trigonometric Double Angle Formula: sin(2θ) = 2sin(θ)cos(θ)

[2sin(x)cos(x)]2 = sin(x)

4sin2(x)cos2(x) = sin(x)

Subtract sin(x) from each side.

4sin2(x)cos2(x) - sin(x) = 0.

Take out common term sin(x)

sin(x)[4sin(x)cos2(x) - 1] = 0

sin(x) = 0 or [4sin(x)cos2(x) - 1] = 0

Take sin(x) = 0

Trigonometric table in sin(0°) =  0.

So, sin(x) = sin(0°) ⇒ x = 0.

Now take [4sin(x)cos2(x) - 1] = 0

Pythagorean Identities: cos2(θ) = 1 - sin2(θ)

4sin(x)[1 - sin2(x)] - 1 = 0

4sin(x) - 4sin3(x) - 1 = 0

Multiply each side by negative one.

4sin3(θ) - 4sin(x) + 1 = 0

Now solve the equation using newton method you get x = 0.996063, x = π = 3.141 and x = 2π = 6.28319.

Therefore x = 0, x = 0.996063, x = π = 3.141 and x = 2π = 6.28319.

answered Mar 7, 2013 by richardson Scholar

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