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

The general solution for 2cos(2x)+sqrt(2)=0 is x=(3pi)/8+pin,x=(5pi)/8+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

2cos(2x)+sqrt(2)=0

Solution

2 cos (2 x) + 2 = 0

Solution

x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn

Degrees

x = 67. 5^{\circ} + 18 0^{\circ} n, x = 112. 5^{\circ} + 18 0^{\circ} n

Solution steps

2 cos (2 x) + 2 = 0

Move

2

to the right side

2 cos (2 x) + 2 = 0

Subtract

2

from both sides

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

Simplify

2 cos (2 x) = - 2

2 cos (2 x) = - 2

Divide both sides by

2

2 cos (2 x) = - 2

Divide both sides by

2

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

Simplify

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

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

General solutions for

cos (2 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}

2 x = \frac{3 π}{4} + 2 πn, 2 x = \frac{5 π}{4} + 2 πn

2 x = \frac{3 π}{4} + 2 πn, 2 x = \frac{5 π}{4} + 2 πn

Solve

2 x = \frac{3 π}{4} + 2 πn : x = \frac{3 π}{8} + 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 = \frac{5 π}{4} + 2 πn : x = \frac{5 π}{8} + πn

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{4} + 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 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{5 π}{4}}{2} + \frac{2 πn}{2} : \frac{5 π}{8} + πn

\frac{\frac{5 π}{4}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{4}}{2} = \frac{5 π}{8}

\frac{\frac{5 π}{4}}{2}

Apply the fraction rule:

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

= \frac{5 π}{4 \cdot 2}

Multiply the numbers:

4 \cdot 2 = 8

= \frac{5 π}{8}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{5 π}{8} + πn

x = \frac{3 π}{8} + πn, x = \frac{5 π}{8} + πn

Popular Examples

5sin(x)cos(x)=-2cos(x)

5 sin (x) cos (x) = - 2 cos (x)

cos(θ)= 1/6

cos (θ) = \frac{1}{6}

2sin(x)-1=csc(x)

2 sin (x) - 1 = csc (x)

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

sin^{s i n (x)} (x) = 2

4cos(x)+2=0

4 cos (x) + 2 = 0

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2x)+sqrt(2)=0 ?
The general solution for 2cos(2x)+sqrt(2)=0 is x=(3pi)/8+pin,x=(5pi)/8+pin