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

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

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

Solution

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

Solution

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

Degrees

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

Solution steps

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

Apply the fraction rule:

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

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

General solutions for

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

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

Solve

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

2 t = \frac{5 π}{6} + 2 πn

Divide both sides by

2

2 t = \frac{5 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

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

t = \frac{5 π}{12} + πn

t = \frac{5 π}{12} + πn

t = \frac{5 π}{12} + πn

Solve

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

2 t = \frac{7 π}{6} + 2 πn

Divide both sides by

2

2 t = \frac{7 π}{6} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 t}{2} : t

\frac{2 t}{2}

Divide the numbers:

\frac{2}{2} = 1

= t

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

t = \frac{7 π}{12} + πn

t = \frac{7 π}{12} + πn

t = \frac{7 π}{12} + πn

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

Popular Examples

cos(2x)=(sqrt(3))/2 ,0<= x<= 2pi

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

arcsin(4/5)

arcsin (\frac{4}{5})

cos(x)= 3/4

cos (x) = \frac{3}{4}

csc(pi/4)

csc (\frac{π}{4})

prove (cos(2x)-cos(4x))/(sin(2x)+sin(4x))=tan(x)prove

\frac{cos ( 2 x ) - cos ( 4 x )}{sin ( 2 x ) + sin ( 4 x )} = tan (x)

Frequently Asked Questions (FAQ)

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