What is the general solution for 0=sqrt(1-cos(2x)) ?

The general solution for 0=sqrt(1-cos(2x)) is x=pin

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0=sqrt(1-cos(2x))

Solution

0 = 1 - cos (2 x)

Solution

x = πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n

Solution steps

0 = 1 - cos (2 x)

Switch sides

1 - cos (2 x) = 0

Square both sides:

1 - cos (2 x) = 0

1 - cos (2 x) = 0

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

Expand

(1 - cos (2 x))^{2} : 1 - cos (2 x)

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

Apply radical rule:

a = a^{\frac{1}{2}}

= ((1 - cos (2 x))^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

= (1 - cos (2 x))^{\frac{1}{2} \cdot 2}

\frac{1}{2} \cdot 2 = 1

\frac{1}{2} \cdot 2

Multiply fractions:

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

= \frac{1 \cdot 2}{2}

Cancel the common factor:

2

= 1

= 1 - cos (2 x)

Expand

0^{2} : 0

0^{2}

Apply rule

0^{a} = 0

= 0

1 - cos (2 x) = 0

1 - cos (2 x) = 0

Solve

1 - cos (2 x) = 0 : cos (2 x) = 1

1 - cos (2 x) = 0

Move

1

to the right side

1 - cos (2 x) = 0

Subtract

1

from both sides

1 - cos (2 x) - 1 = 0 - 1

Simplify

- cos (2 x) = - 1

- cos (2 x) = - 1

Divide both sides by

- 1

- cos (2 x) = - 1

Divide both sides by

- 1

\frac{- cos ( 2 x )}{- 1} = \frac{- 1}{- 1}

Simplify

cos (2 x) = 1

cos (2 x) = 1

cos (2 x) = 1

Verify Solutions:

cos (2 x) = 1

True

Check the solutions by plugging them into

1 - cos (2 x) = 0

Remove the ones that don't agree with the equation.

Plug in

cos (2 x) = 1 :

True

1 - 1 = 0

1 - 1 = 0

1 - 1

Subtract the numbers:

1 - 1 = 0

= 0

Apply rule

0 = 0

= 0

0 = 0

True

The solution is

cos (2 x) = 1

General solutions for

cos (2 x) = 1

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 = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

Popular Examples

sec^2(x)+2=3sec(x)

sec^{2} (x) + 2 = 3 sec (x)

cos(x)-1=-cos(x)

cos (x) - 1 = - cos (x)

2cos^2(x)=3

2 cos^{2} (x) = 3

9tan(2x)-9cot(x)=0

9 tan (2 x) - 9 cot (x) = 0

sin(2x)=2sin^2(x)

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

Frequently Asked Questions (FAQ)

What is the general solution for 0=sqrt(1-cos(2x)) ?
The general solution for 0=sqrt(1-cos(2x)) is x=pin