What is the general solution for cos(2θ)=-(sqrt(3))/2 ?

The general solution for cos(2θ)=-(sqrt(3))/2 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

cos(2θ)=-(sqrt(3))/2

Solution

cos (2 θ) = - \frac{3}{2}

Solution

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

Degrees

θ = 7 5^{\circ} + 18 0^{\circ} n, θ = 10 5^{\circ} + 18 0^{\circ} n

Solution steps

cos (2 θ) = - \frac{3}{2}

General solutions for

cos (2 θ) = - \frac{3}{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 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

2 θ = \frac{5 π}{6} + 2 πn, 2 θ = \frac{7 π}{6} + 2 πn

Solve

2 θ = \frac{5 π}{6} + 2 πn : θ = \frac{5 π}{12} + πn

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{5 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2} : \frac{5 π}{12} + πn

\frac{\frac{5 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

θ = \frac{5 π}{12} + πn

Solve

2 θ = \frac{7 π}{6} + 2 πn : θ = \frac{7 π}{12} + πn

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 θ = \frac{7 π}{6} + 2 πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} = \frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

Simplify

\frac{2 θ}{2} : θ

\frac{2 θ}{2}

Divide the numbers:

\frac{2}{2} = 1

= θ

Simplify

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2} : \frac{7 π}{12} + πn

\frac{\frac{7 π}{6}}{2} + \frac{2 πn}{2}

\frac{\frac{7 π}{6}}{2} = \frac{7 π}{12}

\frac{\frac{7 π}{6}}{2}

Apply the fraction rule:

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

= \frac{7 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{7 π}{12}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{7 π}{12} + πn

θ = \frac{5 π}{12} + πn, θ = \frac{7 π}{12} + πn

Popular Examples

cot^2(135)-sin(210)+5cos^2(225)

cot^{2} (13 5^{\circ}) - sin (21 0^{\circ}) + 5 cos^{2} (22 5^{\circ})

sin(1)

sin (1)

2sin^2(θ)=1-sin(θ)

2 sin^{2} (θ) = 1 - sin (θ)

sin(2x)-sin(x)=0

sin (2 x) - sin (x) = 0

sin(540)

sin (54 0^{\circ})

Frequently Asked Questions (FAQ)

What is the general solution for cos(2θ)=-(sqrt(3))/2 ?
The general solution for cos(2θ)=-(sqrt(3))/2 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin