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

The general solution for cos^2(4x)-sin^2(4x)=0 is x= pi/(16)+(pin)/4 ,x=(3pi}{16}+\frac{pin)/4

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

Solution

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

Solution

x = \frac{π}{16} + \frac{πn}{4}, x = \frac{3 π}{16} + \frac{πn}{4}

Degrees

x = 11.2 5^{\circ} + 4 5^{\circ} n, x = 33.7 5^{\circ} + 4 5^{\circ} n

Solution steps

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

Rewrite using trig identities

cos^{2} (4 x) - sin^{2} (4 x)

Use the Double Angle identity:

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

= cos (2 \cdot 4 x)

cos (2 \cdot 4 x) = 0

General solutions for

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

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

Solve

2 \cdot 4 x = \frac{π}{2} + 2 πn : x = \frac{π}{16} + \frac{πn}{4}

2 \cdot 4 x = \frac{π}{2} + 2 πn

Divide both sides by

8

2 \cdot 4 x = \frac{π}{2} + 2 πn

Divide both sides by

8

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

Simplify

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

Simplify

\frac{2 \cdot 4 x}{8} : x

\frac{2 \cdot 4 x}{8}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8 x}{8}

Divide the numbers:

\frac{8}{8} = 1

= x

Simplify

\frac{\frac{π}{2}}{8} + \frac{2 πn}{8} : \frac{π}{16} + \frac{πn}{4}

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

\frac{\frac{π}{2}}{8} = \frac{π}{16}

\frac{\frac{π}{2}}{8}

Apply the fraction rule:

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

= \frac{π}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{π}{16}

\frac{2 πn}{8} = \frac{πn}{4}

\frac{2 πn}{8}

Cancel the common factor:

2

= \frac{πn}{4}

= \frac{π}{16} + \frac{πn}{4}

x = \frac{π}{16} + \frac{πn}{4}

x = \frac{π}{16} + \frac{πn}{4}

x = \frac{π}{16} + \frac{πn}{4}

Solve

2 \cdot 4 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{16} + \frac{πn}{4}

2 \cdot 4 x = \frac{3 π}{2} + 2 πn

Divide both sides by

8

2 \cdot 4 x = \frac{3 π}{2} + 2 πn

Divide both sides by

8

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

Simplify

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

Simplify

\frac{2 \cdot 4 x}{8} : x

\frac{2 \cdot 4 x}{8}

Multiply the numbers:

2 \cdot 4 = 8

= \frac{8 x}{8}

Divide the numbers:

\frac{8}{8} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{8} + \frac{2 πn}{8} : \frac{3 π}{16} + \frac{πn}{4}

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

\frac{\frac{3 π}{2}}{8} = \frac{3 π}{16}

\frac{\frac{3 π}{2}}{8}

Apply the fraction rule:

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

= \frac{3 π}{2 \cdot 8}

Multiply the numbers:

2 \cdot 8 = 16

= \frac{3 π}{16}

\frac{2 πn}{8} = \frac{πn}{4}

\frac{2 πn}{8}

Cancel the common factor:

2

= \frac{πn}{4}

= \frac{3 π}{16} + \frac{πn}{4}

x = \frac{3 π}{16} + \frac{πn}{4}

x = \frac{3 π}{16} + \frac{πn}{4}

x = \frac{3 π}{16} + \frac{πn}{4}

x = \frac{π}{16} + \frac{πn}{4}, x = \frac{3 π}{16} + \frac{πn}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for cos^2(4x)-sin^2(4x)=0 ?
The general solution for cos^2(4x)-sin^2(4x)=0 is x= pi/(16)+(pin)/4 ,x=(3pi}{16}+\frac{pin)/4