What is the general solution for sin^2(θ)cos(θ)+cos^3(θ)=-1/2 ?

The general solution for sin^2(θ)cos(θ)+cos^3(θ)=-1/2 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin

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sin^2(θ)cos(θ)+cos^3(θ)=-1/2

Solution

sin^{2} (θ) cos (θ) + cos^{3} (θ) = - \frac{1}{2}

Solution

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

Degrees

θ = 12 0^{\circ} + 36 0^{\circ} n, θ = 24 0^{\circ} + 36 0^{\circ} n

Solution steps

sin^{2} (θ) cos (θ) + cos^{3} (θ) = - \frac{1}{2}

Subtract

- \frac{1}{2}

from both sides

sin^{2} (θ) cos (θ) + cos^{3} (θ) + \frac{1}{2} = 0

Simplify

sin^{2} (θ) cos (θ) + cos^{3} (θ) + \frac{1}{2} : \frac{2 sin ^{2} ( θ ) cos ( θ ) + 2 cos ^{3} ( θ ) + 1}{2}

sin^{2} (θ) cos (θ) + cos^{3} (θ) + \frac{1}{2}

Convert element to fraction:

sin^{2} (θ) cos (θ) = \frac{sin ^{2} ( θ ) cos ( θ ) 2}{2}, cos^{3} (θ) = \frac{cos ^{3} ( θ ) 2}{2}

= \frac{sin ^{2} ( θ ) cos ( θ ) \cdot 2}{2} + \frac{cos ^{3} ( θ ) \cdot 2}{2} + \frac{1}{2}

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \frac{sin ^{2} ( θ ) cos ( θ ) \cdot 2 + cos ^{3} ( θ ) \cdot 2 + 1}{2}

\frac{2 sin ^{2} ( θ ) cos ( θ ) + 2 cos ^{3} ( θ ) + 1}{2} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

2 sin^{2} (θ) cos (θ) + 2 cos^{3} (θ) + 1 = 0

Rewrite using trig identities

1 + 2 cos^{3} (θ) + 2 cos (θ) sin^{2} (θ)

Use the Pythagorean identity:

cos^{2} (x) + sin^{2} (x) = 1

sin^{2} (x) = 1 - cos^{2} (x)

= 1 + 2 cos^{3} (θ) + 2 cos (θ) (1 - cos^{2} (θ))

Simplify

1 + 2 cos^{3} (θ) + 2 cos (θ) (1 - cos^{2} (θ)) : 2 cos (θ) + 1

1 + 2 cos^{3} (θ) + 2 cos (θ) (1 - cos^{2} (θ))

Expand

2 cos (θ) (1 - cos^{2} (θ)) : 2 cos (θ) - 2 cos^{3} (θ)

2 cos (θ) (1 - cos^{2} (θ))

Apply the distributive law:

a (b - c) = ab - a c

a = 2 cos (θ), b = 1, c = cos^{2} (θ)

= 2 cos (θ) \cdot 1 - 2 cos (θ) cos^{2} (θ)

= 2 \cdot 1 \cdot cos (θ) - 2 cos^{2} (θ) cos (θ)

Simplify

2 \cdot 1 \cdot cos (θ) - 2 cos^{2} (θ) cos (θ) : 2 cos (θ) - 2 cos^{3} (θ)

2 \cdot 1 \cdot cos (θ) - 2 cos^{2} (θ) cos (θ)

2 \cdot 1 \cdot cos (θ) = 2 cos (θ)

2 \cdot 1 \cdot cos (θ)

Multiply the numbers:

2 \cdot 1 = 2

= 2 cos (θ)

2 cos^{2} (θ) cos (θ) = 2 cos^{3} (θ)

2 cos^{2} (θ) cos (θ)

Apply exponent rule:

a^{b} \cdot a^{c} = a^{b + c}

cos^{2} (θ) cos (θ) = cos^{2 + 1} (θ)

= 2 cos^{2 + 1} (θ)

Add the numbers:

2 + 1 = 3

= 2 cos^{3} (θ)

= 2 cos (θ) - 2 cos^{3} (θ)

= 2 cos (θ) - 2 cos^{3} (θ)

= 1 + 2 cos^{3} (θ) + 2 cos (θ) - 2 cos^{3} (θ)

Simplify

1 + 2 cos^{3} (θ) + 2 cos (θ) - 2 cos^{3} (θ) : 2 cos (θ) + 1

1 + 2 cos^{3} (θ) + 2 cos (θ) - 2 cos^{3} (θ)

Group like terms

= 2 cos^{3} (θ) + 2 cos (θ) - 2 cos^{3} (θ) + 1

Add similar elements:

2 cos^{3} (θ) - 2 cos^{3} (θ) = 0

= 2 cos (θ) + 1

= 2 cos (θ) + 1

= 2 cos (θ) + 1

1 + 2 cos (θ) = 0

Move

1

to the right side

1 + 2 cos (θ) = 0

Subtract

1

from both sides

1 + 2 cos (θ) - 1 = 0 - 1

Simplify

2 cos (θ) = - 1

2 cos (θ) = - 1

Divide both sides by

2

2 cos (θ) = - 1

Divide both sides by

2

\frac{2 cos ( θ )}{2} = \frac{- 1}{2}

Simplify

cos (θ) = - \frac{1}{2}

cos (θ) = - \frac{1}{2}

General solutions for

cos (θ) = - \frac{1}{2}

cos (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

θ = \frac{2 π}{3} + 2 πn, θ = \frac{4 π}{3} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(θ)cos(θ)+cos^3(θ)=-1/2 ?
The general solution for sin^2(θ)cos(θ)+cos^3(θ)=-1/2 is θ=(2pi)/3+2pin,θ=(4pi)/3+2pin