What is the general solution for 3cos^2(A)+3cos(A)=0 ?

The general solution for 3cos^2(A)+3cos(A)=0 is A= pi/2+2pin,A=(3pi)/2+2pin,A=pi+2pin

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3cos^2(A)+3cos(A)=0

Solution

3 cos^{2} (A) + 3 cos (A) = 0

Solution

A = \frac{π}{2} + 2 πn, A = \frac{3 π}{2} + 2 πn, A = π + 2 πn

Degrees

A = 9 0^{\circ} + 36 0^{\circ} n, A = 27 0^{\circ} + 36 0^{\circ} n, A = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

3 cos^{2} (A) + 3 cos (A) = 0

Solve by substitution

3 cos^{2} (A) + 3 cos (A) = 0

Let:

cos (A) = u

3 u^{2} + 3 u = 0

3 u^{2} + 3 u = 0 : u = 0, u = - 1

3 u^{2} + 3 u = 0

Solve with the quadratic formula

3 u^{2} + 3 u = 0

Quadratic Equation Formula:

For

a = 3, b = 3, c = 0

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

u_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 3 \cdot 0}{2 \cdot 3}

3^{2} - 4 \cdot 3 \cdot 0 = 3

3^{2} - 4 \cdot 3 \cdot 0

Apply rule

0 \cdot a = 0

= 3^{2} - 0

3^{2} - 0 = 3^{2}

= 3^{2}

Apply radical rule:

assuming

a \geq 0

= 3

u_{1, 2} = \frac{- 3 \pm 3}{2 \cdot 3}

Separate the solutions

u_{1} = \frac{- 3 + 3}{2 \cdot 3}, u_{2} = \frac{- 3 - 3}{2 \cdot 3}

u = \frac{- 3 + 3}{2 \cdot 3} : 0

\frac{- 3 + 3}{2 \cdot 3}

Add/Subtract the numbers:

- 3 + 3 = 0

= \frac{0}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{0}{6}

Apply rule

\frac{0}{a} = 0, a \neq = 0

= 0

u = \frac{- 3 - 3}{2 \cdot 3} : - 1

\frac{- 3 - 3}{2 \cdot 3}

Subtract the numbers:

- 3 - 3 = - 6

= \frac{- 6}{2 \cdot 3}

Multiply the numbers:

2 \cdot 3 = 6

= \frac{- 6}{6}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{6}{6}

Apply rule

\frac{a}{a} = 1

= - 1

The solutions to the quadratic equation are:

u = 0, u = - 1

Substitute back

u = cos (A)

cos (A) = 0, cos (A) = - 1

cos (A) = 0, cos (A) = - 1

cos (A) = 0 : A = \frac{π}{2} + 2 πn, A = \frac{3 π}{2} + 2 πn

cos (A) = 0

General solutions for

cos (A) = 0

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}

A = \frac{π}{2} + 2 πn, A = \frac{3 π}{2} + 2 πn

A = \frac{π}{2} + 2 πn, A = \frac{3 π}{2} + 2 πn

cos (A) = - 1 : A = π + 2 πn

cos (A) = - 1

General solutions for

cos (A) = - 1

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}

A = π + 2 πn

A = π + 2 πn

Combine all the solutions

A = \frac{π}{2} + 2 πn, A = \frac{3 π}{2} + 2 πn, A = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 3cos^2(A)+3cos(A)=0 ?
The general solution for 3cos^2(A)+3cos(A)=0 is A= pi/2+2pin,A=(3pi)/2+2pin,A=pi+2pin