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

The general solution for 4cos^2(2x)-1=1 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin

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

Solution

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

Solution

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

Degrees

x = 22. 5^{\circ} + 18 0^{\circ} n, x = 157. 5^{\circ} + 18 0^{\circ} n, x = 67. 5^{\circ} + 18 0^{\circ} n, x = - 67. 5^{\circ} + 18 0^{\circ} n

Solution steps

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

Solve by substitution

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

Let:

cos (2 x) = u

4 u^{2} - 1 = 1

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

4 u^{2} - 1 = 1

Move

1

to the right side

4 u^{2} - 1 = 1

Add

1

to both sides

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

Simplify

4 u^{2} = 2

4 u^{2} = 2

Divide both sides by

4

4 u^{2} = 2

Divide both sides by

4

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

Simplify

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

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

For

x^{2} = f (a)

the solutions are

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

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}{8} + πn, x = π - \frac{π}{8} + πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = a \Rightarrow x = arccos (a) + 2 πn, x = 2 π - arccos (a) + 2 πn

2 x = arccos (\frac{1}{2}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{2}) + 2 πn

2 x = arccos (\frac{1}{2}) + 2 πn, 2 x = 2 π - arccos (\frac{1}{2}) + 2 πn

Solve

2 x = arccos (\frac{1}{2}) + 2 πn : x = \frac{π}{8} + πn

2 x = arccos (\frac{1}{2}) + 2 πn

Simplify

arccos (\frac{1}{2}) + 2 πn : \frac{π}{4} + 2 πn

arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{π}{4} + 2 πn

2 x = \frac{π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{π}{4} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{π}{4}}{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}{4}}{2} + \frac{2 πn}{2} : \frac{π}{8} + πn

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

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

x = \frac{π}{8} + πn

Solve

2 x = 2 π - arccos (\frac{1}{2}) + 2 πn : x = π - \frac{π}{8} + πn

2 x = 2 π - arccos (\frac{1}{2}) + 2 πn

Simplify

2 π - arccos (\frac{1}{2}) + 2 πn : 2 π - \frac{π}{4} + 2 πn

2 π - arccos (\frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (\frac{1}{2}) = \frac{π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= 2 π - \frac{π}{4} + 2 πn

2 x = 2 π - \frac{π}{4} + 2 πn

Divide both sides by

2

2 x = 2 π - \frac{π}{4} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{2 π}{2} - \frac{\frac{π}{4}}{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{2 π}{2} - \frac{\frac{π}{4}}{2} + \frac{2 πn}{2} : π - \frac{π}{8} + πn

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

\frac{2 π}{2} = π

\frac{2 π}{2}

Divide the numbers:

\frac{2}{2} = 1

= π

\frac{\frac{π}{4}}{2} = \frac{π}{8}

\frac{\frac{π}{4}}{2}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = π - \frac{π}{8} + πn

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn

cos (2 x) = - \frac{1}{2} : x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

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

Apply trig inverse properties

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

General solutions for

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

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

2 x = arccos (- \frac{1}{2}) + 2 πn, 2 x = - arccos (- \frac{1}{2}) + 2 πn

2 x = arccos (- \frac{1}{2}) + 2 πn, 2 x = - arccos (- \frac{1}{2}) + 2 πn

Solve

2 x = arccos (- \frac{1}{2}) + 2 πn : x = \frac{3 π}{8} + πn

2 x = arccos (- \frac{1}{2}) + 2 πn

Simplify

arccos (- \frac{1}{2}) + 2 πn : \frac{3 π}{4} + 2 πn

arccos (- \frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= \frac{3 π}{4} + 2 πn

2 x = \frac{3 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{3 π}{4} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = \frac{\frac{3 π}{4}}{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{3 π}{4}}{2} + \frac{2 πn}{2} : \frac{3 π}{8} + πn

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn

Solve

2 x = - arccos (- \frac{1}{2}) + 2 πn : x = - \frac{3 π}{8} + πn

2 x = - arccos (- \frac{1}{2}) + 2 πn

Simplify

- arccos (- \frac{1}{2}) + 2 πn : - \frac{3 π}{4} + 2 πn

- arccos (- \frac{1}{2}) + 2 πn

Use the following trivial identity:

arccos (- \frac{1}{2}) = \frac{3 π}{4}

x - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 arccos (x) π \frac{5 π}{6} \frac{3 π}{4} \frac{2 π}{3} \frac{π}{2} \frac{π}{3} \frac{π}{4} \frac{π}{6} 0 arccos (x) 18 0^{\circ} 15 0^{\circ} 13 5^{\circ} 12 0^{\circ} 9 0^{\circ} 6 0^{\circ} 4 5^{\circ} 3 0^{\circ} 0^{\circ}

= - \frac{3 π}{4} + 2 πn

2 x = - \frac{3 π}{4} + 2 πn

Divide both sides by

2

2 x = - \frac{3 π}{4} + 2 πn

Divide both sides by

2

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

Simplify

\frac{2 x}{2} = - \frac{\frac{3 π}{4}}{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{3 π}{4}}{2} + \frac{2 πn}{2} : - \frac{3 π}{8} + πn

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

\frac{\frac{3 π}{4}}{2} = \frac{3 π}{8}

\frac{\frac{3 π}{4}}{2}

Apply the fraction rule:

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

= \frac{3 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{3 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = - \frac{3 π}{8} + πn

x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

Combine all the solutions

x = \frac{π}{8} + πn, x = π - \frac{π}{8} + πn, x = \frac{3 π}{8} + πn, x = - \frac{3 π}{8} + πn

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Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(2x)-1=1 ?
The general solution for 4cos^2(2x)-1=1 is x= pi/8+pin,x=pi-pi/8+pin,x=(3pi)/8+pin,x=-(3pi)/8+pin