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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos^2(2x)= 1/2

Solution

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

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

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

Solve by substitution

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

Let:

cos (2 x) = u

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

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

Popular Examples

sin(x)+2cos^2(x)=1

sin (x) + 2 cos^{2} (x) = 1

7(1-cos(θ))=sin^2(θ)

7 (1 - cos (θ)) = sin^{2} (θ)

2cos(x)-sin^2(x)+2=0

2 cos (x) - sin^{2} (x) + 2 = 0

sinh(x)=2

sinh (x) = 2

cos(2t)= 1/2

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

Frequently Asked Questions (FAQ)

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