What is the general solution for cos(a)+1=4cos(a)+1 ?

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

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cos(a)+1=4cos(a)+1

cos (a) + 1 = 4 cos (a) + 1

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

Degrees

a = 9 0^{\circ} + 36 0^{\circ} n, a = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

cos (a) + 1 = 4 cos (a) + 1

Solve by substitution

cos (a) + 1 = 4 cos (a) + 1

Let:

cos (a) = u

u + 1 = 4 u + 1

u + 1 = 4 u + 1 : u = 0

u + 1 = 4 u + 1

Subtract

1

from both sides

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

Simplify

u = 4 u

Move

4 u

to the left side

u = 4 u

Subtract

4 u

from both sides

u - 4 u = 4 u - 4 u

Simplify

- 3 u = 0

- 3 u = 0

Divide both sides by

- 3

- 3 u = 0

Divide both sides by

- 3

\frac{- 3 u}{- 3} = \frac{0}{- 3}

Simplify

u = 0

u = 0

Substitute back

u = cos (a)

cos (a) = 0

cos (a) = 0

cos (a) = 0 : a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

cos (a) = 0

General solutions for

cos (a) = 0

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}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

Combine all the solutions

a = \frac{π}{2} + 2 πn, a = \frac{3 π}{2} + 2 πn

6 cos (x) + 3 sin (x) = 5

(2 sin (x) - 1) cos (x) = 0

sin (2 x) + cos (x) = 0, x \leq 2 π, 0

6 sin (x) = 5 cos (x)

sin (θ) = \frac{- 3}{4}

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