What is the general solution for cos(x)=(4(-1/2)-1)/3 ?

The general solution for cos(x)=(4(-1/2)-1)/3 is x=pi+2pin

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cos(x)=(4(-1/2)-1)/3

cos (x) = \frac{4 ( - \frac{1}{2} ) - 1}{3}

x = π + 2 πn

Solution steps

cos (x) = \frac{4 ( - \frac{1}{2} ) - 1}{3}

Simplify

\frac{4 ( - \frac{1}{2} ) - 1}{3} : - 1

\frac{4 ( - \frac{1}{2} ) - 1}{3}

Remove parentheses:

(- a) = - a

= \frac{- 4 \cdot \frac{1}{2} - 1}{3}

4 \cdot \frac{1}{2} = 2

4 \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{1 \cdot 4}{2}

Multiply the numbers:

1 \cdot 4 = 4

= \frac{4}{2}

Divide the numbers:

\frac{4}{2} = 2

= 2

= \frac{- 2 - 1}{3}

Subtract the numbers:

- 2 - 1 = - 3

= \frac{- 3}{3}

Apply the fraction rule:

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

= - \frac{3}{3}

Apply rule

\frac{a}{a} = 1

= - 1

General solutions for

cos (x) = - 1

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}

x = π + 2 πn

x = π + 2 πn

1 = l sin (x), 2 = l cos (x)

y, t = 2 cot (6 0^{\circ} y + 3 0^{\circ})

sec (y) = - 2

- 1 + 2 sin^{2} (x) = 0

θ, y = - 2 + cos (3 θ)

What is the general solution for cos(x)=(4(-1/2)-1)/3 ?
The general solution for cos(x)=(4(-1/2)-1)/3 is x=pi+2pin