What is the general solution for (2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1 ?

The general solution for (2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1 is a=2pin,a=pi+2pin

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(2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1

Solution

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

Solution

a = 2 πn, a = π + 2 πn

Degrees

a = 0^{\circ} + 36 0^{\circ} n, a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

Solve by substitution

(2 cos (a) + 1) (2 cos (a) - 1) = 2 cos^{2} (a) + 1

Let:

cos (a) = u

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

(2 u + 1) (2 u - 1) = 2 u^{2} + 1 : u = 1, u = - 1

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

Expand

(2 u + 1) (2 u - 1) : 4 u^{2} - 1

(2 u + 1) (2 u - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 2 u, b = 1

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

Simplify

(2 u)^{2} - 1^{2} : 4 u^{2} - 1

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

Apply rule

1^{a} = 1

1^{2} = 1

= (2 u)^{2} - 1

(2 u)^{2} = 4 u^{2}

(2 u)^{2}

Apply exponent rule:

(a \cdot b)^{n} = a^{n} b^{n}

= 2^{2} u^{2}

2^{2} = 4

= 4 u^{2}

= 4 u^{2} - 1

= 4 u^{2} - 1

4 u^{2} - 1 = 2 u^{2} + 1

Move

1

to the right side

4 u^{2} - 1 = 2 u^{2} + 1

Add

1

to both sides

4 u^{2} - 1 + 1 = 2 u^{2} + 1 + 1

Simplify

4 u^{2} = 2 u^{2} + 2

4 u^{2} = 2 u^{2} + 2

Move

2 u^{2}

to the left side

4 u^{2} = 2 u^{2} + 2

Subtract

2 u^{2}

from both sides

4 u^{2} - 2 u^{2} = 2 u^{2} + 2 - 2 u^{2}

Simplify

2 u^{2} = 2

2 u^{2} = 2

Divide both sides by

2

2 u^{2} = 2

Divide both sides by

2

\frac{2 u ^{2}}{2} = \frac{2}{2}

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cos (a)

cos (a) = 1, cos (a) = - 1

cos (a) = 1, cos (a) = - 1

cos (a) = 1 : a = 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 = 0 + 2 πn

a = 0 + 2 πn

Solve

a = 0 + 2 πn : a = 2 πn

a = 0 + 2 πn

0 + 2 πn = 2 πn

a = 2 πn

a = 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 = 2 πn, a = π + 2 πn

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Frequently Asked Questions (FAQ)

What is the general solution for (2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1 ?
The general solution for (2cos(a)+1)(2cos(a)-1)=2cos^2(a)+1 is a=2pin,a=pi+2pin