What is the general solution for 4cos(A)+3=2cos(A)+4 ?

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

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4cos(A)+3=2cos(A)+4

4 cos (A) + 3 = 2 cos (A) + 4

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

4 cos (A) + 3 = 2 cos (A) + 4

Solve by substitution

4 cos (A) + 3 = 2 cos (A) + 4

Let:

cos (A) = u

4 u + 3 = 2 u + 4

4 u + 3 = 2 u + 4 : u = \frac{1}{2}

4 u + 3 = 2 u + 4

Move

3

to the right side

4 u + 3 = 2 u + 4

Subtract

3

from both sides

4 u + 3 - 3 = 2 u + 4 - 3

Simplify

4 u = 2 u + 1

4 u = 2 u + 1

Move

2 u

to the left side

4 u = 2 u + 1

Subtract

2 u

from both sides

4 u - 2 u = 2 u + 1 - 2 u

Simplify

2 u = 1

2 u = 1

Divide both sides by

2

2 u = 1

Divide both sides by

2

\frac{2 u}{2} = \frac{1}{2}

Simplify

u = \frac{1}{2}

u = \frac{1}{2}

Substitute back

u = cos (A)

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

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

cos (A) = \frac{1}{2} : A = \frac{π}{3} + 2 πn, A = \frac{5 π}{3} + 2 πn

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

Combine all the solutions

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

sin^{2} (x) - tan (x) = 0

6 sin^{3} (x) + 7 sin^{2} (x) - sin (x) - 2 = 0

2 sin (x) cos^{2} (x) - sin (x) = 0

tan (x) = \frac{1}{sec ( x )}

2 \cdot cos (2 x) - 1 = 0

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