What is the general solution for -2cos(a)+6=cos(a)+9 ?

The general solution for -2cos(a)+6=cos(a)+9 is a=pi+2pin

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-2cos(a)+6=cos(a)+9

- 2 cos (a) + 6 = cos (a) + 9

a = π + 2 πn

Degrees

a = 18 0^{\circ} + 36 0^{\circ} n

Solution steps

- 2 cos (a) + 6 = cos (a) + 9

Solve by substitution

- 2 cos (a) + 6 = cos (a) + 9

Let:

cos (a) = u

- 2 u + 6 = u + 9

- 2 u + 6 = u + 9 : u = - 1

- 2 u + 6 = u + 9

Move

6

to the right side

- 2 u + 6 = u + 9

Subtract

6

from both sides

- 2 u + 6 - 6 = u + 9 - 6

Simplify

- 2 u = u + 3

- 2 u = u + 3

Move

u

to the left side

- 2 u = u + 3

Subtract

u

from both sides

- 2 u - u = u + 3 - u

Simplify

- 3 u = 3

- 3 u = 3

Divide both sides by

- 3

- 3 u = 3

Divide both sides by

- 3

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

Simplify

u = - 1

u = - 1

Substitute back

u = cos (a)

cos (a) = - 1

cos (a) = - 1

cos (a) = - 1 : a = π + 2 πn

cos (a) = - 1

General solutions for

cos (a) = - 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}

a = π + 2 πn

a = π + 2 πn

Combine all the solutions

a = π + 2 πn

sin (θ) = 0.64

tan (θ) = \frac{6}{5}

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

y, x = arccos (y)

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

What is the general solution for -2cos(a)+6=cos(a)+9 ?
The general solution for -2cos(a)+6=cos(a)+9 is a=pi+2pin