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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

4cos^2(θ)=1,0<= θ<2pi

Solution

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

Solution

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

Degrees

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

Solution steps

4 cos^{2} (θ) = 1, 0 \leq θ < 2 π

Solve by substitution

4 cos^{2} (θ) = 1

Let:

cos (θ) = u

4 u^{2} = 1

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

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 (θ)

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

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

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

cos (θ) = \frac{1}{2}, 0 \leq θ < 2 π

General solutions for

cos (θ) = \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}

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

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{3}, θ = \frac{5 π}{3}

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

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

General solutions for

cos (θ) = - \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}

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

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{2 π}{3}, θ = \frac{4 π}{3}

Combine all the solutions

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

Popular Examples

cos(x/2+pi/3)=(sqrt(2))/2

cos (\frac{x}{2} + \frac{π}{3}) = \frac{2}{2}

tan(θ)=8.2

tan (θ) = 8.2

2sin(x)=4sin(x)+1

2 sin (x) = 4 sin (x) + 1

-1-cos(θ)=-3cos(θ)

- 1 - cos (θ) = - 3 cos (θ)

3*cos(θ)=2-sin(θ)

3 \cdot cos (θ) = 2 - sin (θ)

Frequently Asked Questions (FAQ)

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