What is the general solution for 2cos^2(4x+1)=0 ?

The general solution for 2cos^2(4x+1)=0 is x=-1/4+pi/8+(pin)/2 ,x=-1/4+(3pi)/8+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos^2(4x+1)=0

Solution

2 cos^{2} (4 x + 1) = 0

Solution

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

Degrees

x = 8.17605 \dots^{\circ} + 9 0^{\circ} n, x = 53.17605 \dots^{\circ} + 9 0^{\circ} n

Solution steps

2 cos^{2} (4 x + 1) = 0

Divide both sides by

2

2 cos^{2} (4 x + 1) = 0

Divide both sides by

2

2 cos^{2} (4 x + 1) = 0

Divide both sides by

2

\frac{2 cos ^{2} ( 4 x + 1 )}{2} = \frac{0}{2}

Simplify

cos^{2} (4 x + 1) = 0

cos^{2} (4 x + 1) = 0

Apply rule

x^{n} = 0 \Rightarrow x = 0

cos (4 x + 1) = 0

General solutions for

cos (4 x + 1) = 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}

4 x + 1 = \frac{π}{2} + 2 πn, 4 x + 1 = \frac{3 π}{2} + 2 πn

4 x + 1 = \frac{π}{2} + 2 πn, 4 x + 1 = \frac{3 π}{2} + 2 πn

Solve

4 x + 1 = \frac{π}{2} + 2 πn : x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

4 x + 1 = \frac{π}{2} + 2 πn

Move

1

to the right side

4 x + 1 = \frac{π}{2} + 2 πn

Subtract

1

from both sides

4 x + 1 - 1 = \frac{π}{2} + 2 πn - 1

Simplify

4 x = \frac{π}{2} + 2 πn - 1

4 x = \frac{π}{2} + 2 πn - 1

Divide both sides by

4

4 x = \frac{π}{2} + 2 πn - 1

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4} : - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

\frac{\frac{π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Group like terms

= - \frac{1}{4} + \frac{2 πn}{4} + \frac{\frac{π}{2}}{4}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

\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{1}{4} + \frac{πn}{2} + \frac{π}{8}

Group like terms

= - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}

Solve

4 x + 1 = \frac{3 π}{2} + 2 πn : x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

4 x + 1 = \frac{3 π}{2} + 2 πn

Move

1

to the right side

4 x + 1 = \frac{3 π}{2} + 2 πn

Subtract

1

from both sides

4 x + 1 - 1 = \frac{3 π}{2} + 2 πn - 1

Simplify

4 x = \frac{3 π}{2} + 2 πn - 1

4 x = \frac{3 π}{2} + 2 πn - 1

Divide both sides by

4

4 x = \frac{3 π}{2} + 2 πn - 1

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4} : - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

\frac{\frac{3 π}{2}}{4} + \frac{2 πn}{4} - \frac{1}{4}

Group like terms

= - \frac{1}{4} + \frac{2 πn}{4} + \frac{\frac{3 π}{2}}{4}

\frac{2 πn}{4} = \frac{πn}{2}

\frac{2 πn}{4}

Cancel the common factor:

2

= \frac{πn}{2}

\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{1}{4} + \frac{πn}{2} + \frac{3 π}{8}

Group like terms

= - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

x = - \frac{1}{4} + \frac{π}{8} + \frac{πn}{2}, x = - \frac{1}{4} + \frac{3 π}{8} + \frac{πn}{2}

Popular Examples

solvefor x,2sin(x)=-1 solve for

x, 2 sin (x) = - 1

solvefor t,f=acos((tpi)/6-(11pi)/(12))solve for

t, f = a cos (\frac{t π}{6} - \frac{11 π}{12})

-9cos^2(θ)+9=16sin(θ)-7,0<= θ<360

- 9 cos^{2} (θ) + 9 = 16 sin (θ) - 7, 0 \leq θ < 36 0^{\circ}

(cos(x))/(sin^2(x))= 2/3

\frac{cos ( x )}{sin ^{2} ( x )} = \frac{2}{3}

7cos(b)-sqrt(13)=0

7 cos (b) - 13 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cos^2(4x+1)=0 ?
The general solution for 2cos^2(4x+1)=0 is x=-1/4+pi/8+(pin)/2 ,x=-1/4+(3pi)/8+(pin)/2