What is the general solution for 4cos^2(2x)-1=0 ?

The general solution for 4cos^2(2x)-1=0 is x= pi/6+pin,x=(5pi)/6+pin,x= pi/3+pin,x=(2pi)/3+pin

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4cos^2(2x)-1=0

Solution

4 cos^{2} (2 x) - 1 = 0

Solution

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

Degrees

x = 3 0^{\circ} + 18 0^{\circ} n, x = 15 0^{\circ} + 18 0^{\circ} n, x = 6 0^{\circ} + 18 0^{\circ} n, x = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

4 cos^{2} (2 x) - 1 = 0

Solve by substitution

4 cos^{2} (2 x) - 1 = 0

Let:

cos (2 x) = u

4 u^{2} - 1 = 0

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

4 u^{2} - 1 = 0

Move

1

to the right side

4 u^{2} - 1 = 0

Add

1

to both sides

4 u^{2} - 1 + 1 = 0 + 1

Simplify

4 u^{2} = 1

4 u^{2} = 1

Divide both sides by

4

4 u^{2} = 1

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{1}{4}

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

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

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

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

Substitute back

u = cos (2 x)

cos (2 x) = \frac{1}{2}, cos (2 x) = - \frac{1}{2}

cos (2 x) = \frac{1}{2}, cos (2 x) = - \frac{1}{2}

cos (2 x) = \frac{1}{2} : x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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

General solutions for

cos (2 x) = \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}

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

2 x = \frac{5 π}{3} + 2 πn : x = \frac{5 π}{6} + πn

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2} : \frac{5 π}{6} + πn

\frac{\frac{5 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{3}}{2} = \frac{5 π}{6}

\frac{\frac{5 π}{3}}{2}

Apply the fraction rule:

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

= \frac{5 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{5 π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn

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

cos (2 x) = - \frac{1}{2}

General solutions for

cos (2 x) = - \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}

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

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

Solve

2 x = \frac{2 π}{3} + 2 πn : x = \frac{π}{3} + πn

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{3} + πn

\frac{\frac{2 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

Solve

2 x = \frac{4 π}{3} + 2 πn : x = \frac{2 π}{3} + πn

2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{4 π}{3} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2} : \frac{2 π}{3} + πn

\frac{\frac{4 π}{3}}{2} + \frac{2 πn}{2}

\frac{\frac{4 π}{3}}{2} = \frac{2 π}{3}

\frac{\frac{4 π}{3}}{2}

Apply the fraction rule:

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

= \frac{4 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{4 π}{6}

Cancel the common factor:

2

= \frac{2 π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

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

Combine all the solutions

x = \frac{π}{6} + πn, x = \frac{5 π}{6} + πn, x = \frac{π}{3} + πn, x = \frac{2 π}{3} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(2x)-1=0 ?
The general solution for 4cos^2(2x)-1=0 is x= pi/6+pin,x=(5pi)/6+pin,x= pi/3+pin,x=(2pi)/3+pin