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

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

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cos(b)+4=3cos(b)+4

cos (b) + 4 = 3 cos (b) + 4

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

Degrees

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

Solution steps

cos (b) + 4 = 3 cos (b) + 4

Solve by substitution

cos (b) + 4 = 3 cos (b) + 4

Let:

cos (b) = u

u + 4 = 3 u + 4

u + 4 = 3 u + 4 : u = 0

u + 4 = 3 u + 4

Subtract

4

from both sides

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

Simplify

u = 3 u

Move

3 u

to the left side

u = 3 u

Subtract

3 u

from both sides

u - 3 u = 3 u - 3 u

Simplify

- 2 u = 0

- 2 u = 0

Divide both sides by

- 2

- 2 u = 0

Divide both sides by

- 2

\frac{- 2 u}{- 2} = \frac{0}{- 2}

Simplify

u = 0

u = 0

Substitute back

u = cos (b)

cos (b) = 0

cos (b) = 0

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

cos (b) = 0

General solutions for

cos (b) = 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}

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

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

Combine all the solutions

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

- 3 cos (2 θ) - 15 sin (θ) + 8 = - 8 sin (θ) + 3

cos (z) = 4

sin (x) = 8

tan (y) = 0

sin (θ) = \frac{6}{10}

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