What is the general solution for 2cos(2θ)+sqrt(3)=0 ?

The general solution for 2cos(2θ)+sqrt(3)=0 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 >

2cos(2θ)+sqrt(3)=0

Solution

2 cos (2 θ) + 3 = 0

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

2 cos (2 θ) + 3 = 0

Move

3

to the right side

2 cos (2 θ) + 3 = 0

Subtract

3

from both sides

2 cos (2 θ) + 3 - 3 = 0 - 3

Simplify

2 cos (2 θ) = - 3

2 cos (2 θ) = - 3

Divide both sides by

2

2 cos (2 θ) = - 3

Divide both sides by

2

\frac{2 cos ( 2 θ )}{2} = \frac{- 3}{2}

Simplify

cos (2 θ) = - \frac{3}{2}

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

sqrt(3)sin(x)+sin(2x)=0

3 sin (x) + sin (2 x) = 0

18sin^2(x)+27sin(x)+9=0

18 sin^{2} (x) + 27 sin (x) + 9 = 0

3sin(θ)+2sin^2(θ)=-1

3 sin (θ) + 2 sin^{2} (θ) = - 1

cot^2(x)+3csc(x)=-3

cot^{2} (x) + 3 csc (x) = - 3

sec(θ)=3

sec (θ) = 3

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(2θ)+sqrt(3)=0 ?
The general solution for 2cos(2θ)+sqrt(3)=0 is θ=(5pi)/(12)+pin,θ=(7pi)/(12)+pin