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

The general solution for cos^2(2x)-sin^2(2x)=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

Solution

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

Solution

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

Degrees

x = 22. 5^{\circ} + 9 0^{\circ} n, x = 67. 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Rewrite using trig identities

cos^{2} (2 x) - sin^{2} (2 x)

Use the Double Angle identity:

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

= cos (2 \cdot 2 x)

cos (2 \cdot 2 x) = 0

General solutions for

cos (2 \cdot 2 x) = 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}

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

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

Solve

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

2 \cdot 2 x = \frac{π}{2} + 2 πn

Divide both sides by

4

2 \cdot 2 x = \frac{π}{2} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

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

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

\frac{\frac{π}{2}}{4} = \frac{π}{8}

\frac{\frac{π}{2}}{4}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

x = \frac{π}{8} + \frac{πn}{2}

Solve

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

2 \cdot 2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

4

2 \cdot 2 x = \frac{3 π}{2} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{2 \cdot 2 x}{4} : x

\frac{2 \cdot 2 x}{4}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

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

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

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

\frac{\frac{3 π}{2}}{4}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 4}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{3 π}{8}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

x = \frac{3 π}{8} + \frac{πn}{2}

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

Popular Examples

prove tan(θ)+cot(θ)=sec(θ)csc(θ)prove

tan (θ) + cot (θ) = sec (θ) csc (θ)

sin(2x)=-(sqrt(3))/2

sin (2 x) = - \frac{3}{2}

sin(2x)= 1/2

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

arctan(4)

arctan (4)

cos(pi/3)

cos (\frac{π}{3})

Frequently Asked Questions (FAQ)

What is the general solution for cos^2(2x)-sin^2(2x)=0 ?
The general solution for cos^2(2x)-sin^2(2x)=0 is x= pi/8+(pin)/2 ,x=(3pi)/8+(pin)/2