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Solve the following for 0 <_ θ <_ 4pi?

0 votes
The <_ represents less than or equal to.

Sin^2[θ] == -Cos[θ]

Would like the solutions to this question. Thanks.
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asked Dec 28, 2012 in TRIGONOMETRY by skylar Apprentice

1 Answer

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Sin^2[θ] = -Cos[θ]

Note : {Sin^2[θ] + Cos^2[θ] = 1 Than Sin^2[θ] = 1 - Cos^2[θ] }

There fore

1 - Cos^2[θ] = - Cos[θ]

Add Cos^2[θ] to each side

1 - Cos^2[θ] + Cos^2[θ] = - Cos[θ] + Cos^2[θ]

Simplify

1 = Cos^2[θ] - Cos[θ]

Subtract 1 from each side.

1 - 1 = Cos^2[θ] - Cos[θ] - 1

Simplify

0 = Cos^2[θ] - Cos[θ] - 1

Cos^2[θ] - Cos[θ] - 1 = 0

Let Cos[θ] = y

Than

y^2 - y - 1 = 0

Formula : y = [ - b + sqrt(b^2-4ac)] / (2a)

Compire the eqations a(x^2) + bx + c = 0

Than a = 1 , b = - 1 , c = - 1

Substitute a = 1 , b = -1 and c = -1 in the formula

y = [ -(-1) + sqrt((-1)^2 - 4(1)(-1))] / 2(1)

Simplify

y = [ 1 + sqrt(1+4)] / 2

y = [ 1 + sqrt(5)] / 2

There fore

y = [ 1 + sqrt(5)] / 2   or   y = [ 1 - sqrt(5)] / 2

but y = Cos[θ]

Cos[θ] = [ 1 + sqrt(5)] / 2  or  Cos[θ] = [ 1 - sqrt(5)] / 2

answered Jan 17, 2013 by richardson Scholar

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