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

The general solution for 2cos(4x)-sqrt(2)=0 is x= pi/(16)+(pin)/2 ,x=(7pi}{16}+\frac{pin)/2

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2cos(4x)-sqrt(2)=0

Solution

2 cos (4 x) - 2 = 0

Solution

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{7 π}{16} + \frac{πn}{2}

Degrees

x = 11.2 5^{\circ} + 9 0^{\circ} n, x = 78.7 5^{\circ} + 9 0^{\circ} n

Solution steps

2 cos (4 x) - 2 = 0

Move

2

to the right side

2 cos (4 x) - 2 = 0

Add

2

to both sides

2 cos (4 x) - 2 + 2 = 0 + 2

Simplify

2 cos (4 x) = 2

2 cos (4 x) = 2

Divide both sides by

2

2 cos (4 x) = 2

Divide both sides by

2

\frac{2 cos ( 4 x )}{2} = \frac{2}{2}

Simplify

cos (4 x) = \frac{2}{2}

cos (4 x) = \frac{2}{2}

General solutions for

cos (4 x) = \frac{2}{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}

4 x = \frac{π}{4} + 2 πn, 4 x = \frac{7 π}{4} + 2 πn

4 x = \frac{π}{4} + 2 πn, 4 x = \frac{7 π}{4} + 2 πn

Solve

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

4 x = \frac{π}{4} + 2 πn

Divide both sides by

4

4 x = \frac{π}{4} + 2 πn

Divide both sides by

4

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

Simplify

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

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

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

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

\frac{\frac{π}{4}}{4} = \frac{π}{16}

\frac{\frac{π}{4}}{4}

Apply the fraction rule:

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

= \frac{π}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}

Solve

4 x = \frac{7 π}{4} + 2 πn : x = \frac{7 π}{16} + \frac{πn}{2}

4 x = \frac{7 π}{4} + 2 πn

Divide both sides by

4

4 x = \frac{7 π}{4} + 2 πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4} : \frac{7 π}{16} + \frac{πn}{2}

\frac{\frac{7 π}{4}}{4} + \frac{2 πn}{4}

\frac{\frac{7 π}{4}}{4} = \frac{7 π}{16}

\frac{\frac{7 π}{4}}{4}

Apply the fraction rule:

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

= \frac{7 π}{4 \cdot 4}

Multiply the numbers:

4 \cdot 4 = 16

= \frac{7 π}{16}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

= \frac{7 π}{16} + \frac{πn}{2}

x = \frac{7 π}{16} + \frac{πn}{2}

x = \frac{7 π}{16} + \frac{πn}{2}

x = \frac{7 π}{16} + \frac{πn}{2}

x = \frac{π}{16} + \frac{πn}{2}, x = \frac{7 π}{16} + \frac{πn}{2}

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Frequently Asked Questions (FAQ)

What is the general solution for 2cos(4x)-sqrt(2)=0 ?
The general solution for 2cos(4x)-sqrt(2)=0 is x= pi/(16)+(pin)/2 ,x=(7pi}{16}+\frac{pin)/2