What is the general solution for 3-4cos(A)=1 ?

The general solution for 3-4cos(A)=1 is A= pi/3+2pin,A=(5pi)/3+2pin

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3-4cos(A)=1

3 - 4 cos (A) = 1

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

Degrees

A = 6 0^{\circ} + 36 0^{\circ} n, A = 30 0^{\circ} + 36 0^{\circ} n

Solution steps

3 - 4 cos (A) = 1

Move

3

to the right side

3 - 4 cos (A) = 1

Subtract

3

from both sides

3 - 4 cos (A) - 3 = 1 - 3

Simplify

- 4 cos (A) = - 2

- 4 cos (A) = - 2

Divide both sides by

- 4

- 4 cos (A) = - 2

Divide both sides by

- 4

\frac{- 4 cos ( A )}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 cos ( A )}{- 4} = \frac{- 2}{- 4}

Simplify

\frac{- 4 cos ( A )}{- 4} : cos (A)

\frac{- 4 cos ( A )}{- 4}

Apply the fraction rule:

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

= \frac{4 cos ( A )}{4}

Divide the numbers:

\frac{4}{4} = 1

= cos (A)

Simplify

\frac{- 2}{- 4} : \frac{1}{2}

\frac{- 2}{- 4}

Apply the fraction rule:

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

= \frac{2}{4}

Cancel the common factor:

2

= \frac{1}{2}

cos (A) = \frac{1}{2}

cos (A) = \frac{1}{2}

cos (A) = \frac{1}{2}

General solutions for

cos (A) = \frac{1}{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}

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

3 tan (x) - 2 cos (x) = 0

sin (a) = - \frac{3}{5}

cot (x) = cot (3 x - 50)

sinh (x) = - 6

tan (θ) = \frac{2}{2}

What is the general solution for 3-4cos(A)=1 ?
The general solution for 3-4cos(A)=1 is A= pi/3+2pin,A=(5pi)/3+2pin