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

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

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cos(2x)=(-sqrt(3))/2

Solution

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

Solution

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

Degrees

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

Solution steps

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

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

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

General solutions for

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

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

Solve

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

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{5 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

x = \frac{5 π}{12} + πn

Solve

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

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 x = \frac{7 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

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

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

x = \frac{7 π}{12} + πn

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

Popular Examples

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

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

tan(-855)

tan (- 85 5^{\circ})

csc(θ)+2=0

csc (θ) + 2 = 0

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

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

identity cos^2(x)identity

cos^{2} (x)

Frequently Asked Questions (FAQ)

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