What is the general solution for 4cos^2(x)-1=0,0<= x<= 2pi ?

Q: What is the general solution for 4cos^2(x)-1=0,0<= x<= 2pi ?

The general solution for 4cos^2(x)-1=0,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=(2pi)/3 ,x=(4pi)/3

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos^2(x)-1=0,0<= x<= 2pi

Solution

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

Solution

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

Degrees

x = 6 0^{\circ}, x = 30 0^{\circ}, x = 12 0^{\circ}, x = 24 0^{\circ}

Solution steps

4 cos^{2} (x) - 1 = 0, 0 \leq x \leq 2 π

Solve by substitution

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

Let:

cos (x) = u

4 u^{2} - 1 = 0

4 u^{2} - 1 = 0 : u = \frac{1}{2}, u = - \frac{1}{2}

4 u^{2} - 1 = 0

Move

1

to the right side

4 u^{2} - 1 = 0

Add

1

to both sides

4 u^{2} - 1 + 1 = 0 + 1

Simplify

4 u^{2} = 1

4 u^{2} = 1

Divide both sides by

4

4 u^{2} = 1

Divide both sides by

4

\frac{4 u ^{2}}{4} = \frac{1}{4}

Simplify

u^{2} = \frac{1}{4}

u^{2} = \frac{1}{4}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = \frac{1}{4}, u = - \frac{1}{4}

\frac{1}{4} = \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

Apply rule

1 = 1

= \frac{1}{2}

- \frac{1}{4} = - \frac{1}{2}

- \frac{1}{4}

Simplify

\frac{1}{4} : \frac{1}{2}

\frac{1}{4}

Apply radical rule:

assuming

a \geq 0, b \geq 0

= \frac{1}{4}

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= \frac{1}{2}

= - \frac{1}{2}

Apply rule

1 = 1

= - \frac{1}{2}

u = \frac{1}{2}, u = - \frac{1}{2}

Substitute back

u = cos (x)

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

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

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

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

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{3}, x = \frac{5 π}{3}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π : x = \frac{2 π}{3}, x = \frac{4 π}{3}

cos (x) = - \frac{1}{2}, 0 \leq x \leq 2 π

General solutions for

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

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

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

Solutions for the range

0 \leq x \leq 2 π

x = \frac{2 π}{3}, x = \frac{4 π}{3}

Combine all the solutions

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

Popular Examples

cos(2x)+sqrt(3)sin(x)=1

cos (2 x) + 3 sin (x) = 1

tan(2θ)-2cos(θ)=0

tan (2 θ) - 2 cos (θ) = 0

tan(θ)= 12/5 ,pi<θ<(3pi)/2

tan (θ) = \frac{12}{5}, π < θ < \frac{3 π}{2}

sqrt(3)tan(3θ)+1=0,0<= θ<= 2pi

3 tan (3 θ) + 1 = 0, 0 \leq θ \leq 2 π

cos(2x)-sqrt(3)sin(x)=1

cos (2 x) - 3 sin (x) = 1

Frequently Asked Questions (FAQ)

What is the general solution for 4cos^2(x)-1=0,0<= x<= 2pi ?
The general solution for 4cos^2(x)-1=0,0<= x<= 2pi is x= pi/3 ,x=(5pi)/3 ,x=(2pi)/3 ,x=(4pi)/3