What is the general solution for 2cos^2(a)=cos(a) ?

The general solution for 2cos^2(a)=cos(a) is a= pi/3+2pin,a=(5pi)/3+2pin,a= pi/2+2pin,a=(3pi)/2+2pin

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2cos^2(a)=cos(a)

Solution

2 cos^{2} (a) = cos (a)

Solution

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

Degrees

a = 6 0^{\circ} + 36 0^{\circ} n, a = 30 0^{\circ} + 36 0^{\circ} n, a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos^{2} (a) = cos (a)

Solve by substitution

2 cos^{2} (a) = cos (a)

Let:

cos (a) = u

2 u^{2} = u

2 u^{2} = u : u = \frac{1}{2}, u = 0

2 u^{2} = u

Move

u

to the left side

2 u^{2} = u

Subtract

u

from both sides

2 u^{2} - u = u - u

Simplify

2 u^{2} - u = 0

2 u^{2} - u = 0

Solve with the quadratic formula

2 u^{2} - u = 0

Quadratic Equation Formula:

For

a = 2, b = - 1, c = 0

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

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

(- 1)^{2} - 4 \cdot 2 \cdot 0 = 1

(- 1)^{2} - 4 \cdot 2 \cdot 0

(- 1)^{2} = 1

(- 1)^{2}

Apply exponent rule:

(- a)^{n} = a^{n},

n

is even

(- 1)^{2} = 1^{2}

= 1^{2}

Apply rule

1^{a} = 1

= 1

4 \cdot 2 \cdot 0 = 0

4 \cdot 2 \cdot 0

Apply rule

0 \cdot a = 0

= 0

= 1 - 0

Subtract the numbers:

1 - 0 = 1

= 1

Apply rule

1 = 1

= 1

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

Separate the solutions

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

u = \frac{- ( - 1 ) + 1}{2 \cdot 2} : \frac{1}{2}

\frac{- ( - 1 ) + 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 + 1}{2 \cdot 2}

Add the numbers:

1 + 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

u = \frac{- ( - 1 ) - 1}{2 \cdot 2} : 0

\frac{- ( - 1 ) - 1}{2 \cdot 2}

Apply rule

- (- a) = a

= \frac{1 - 1}{2 \cdot 2}

Subtract the numbers:

1 - 1 = 0

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{0}{4}

Apply rule

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

= 0

The solutions to the quadratic equation are:

u = \frac{1}{2}, u = 0

Substitute back

u = cos (a)

cos (a) = \frac{1}{2}, cos (a) = 0

cos (a) = \frac{1}{2}, cos (a) = 0

cos (a) = \frac{1}{2} : a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

cos (a) = \frac{1}{2}

General solutions for

cos (a) = \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}

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn

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

Combine all the solutions

a = \frac{π}{3} + 2 πn, a = \frac{5 π}{3} + 2 πn, a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(a)=cos(a) ?
The general solution for 2cos^2(a)=cos(a) is a= pi/3+2pin,a=(5pi)/3+2pin,a= pi/2+2pin,a=(3pi)/2+2pin