What is the general solution for 10cos(c)+1=cos(c)-1 ?

The general solution for 10cos(c)+1=cos(c)-1 is c=1.79488…+2pin,c=-1.79488…+2pin

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10cos(c)+1=cos(c)-1

10 cos (c) + 1 = cos (c) - 1

c = 1.79488 \dots + 2 πn, c = - 1.79488 \dots + 2 πn

Degrees

c = 102.83958 \dots^{\circ} + 36 0^{\circ} n, c = - 102.83958 \dots^{\circ} + 36 0^{\circ} n

Solution steps

10 cos (c) + 1 = cos (c) - 1

Solve by substitution

10 cos (c) + 1 = cos (c) - 1

Let:

cos (c) = u

10 u + 1 = u - 1

10 u + 1 = u - 1 : u = - \frac{2}{9}

10 u + 1 = u - 1

Move

1

to the right side

10 u + 1 = u - 1

Subtract

1

from both sides

10 u + 1 - 1 = u - 1 - 1

Simplify

10 u = u - 2

10 u = u - 2

Move

u

to the left side

10 u = u - 2

Subtract

u

from both sides

10 u - u = u - 2 - u

Simplify

9 u = - 2

9 u = - 2

Divide both sides by

9

9 u = - 2

Divide both sides by

9

\frac{9 u}{9} = \frac{- 2}{9}

Simplify

u = - \frac{2}{9}

u = - \frac{2}{9}

Substitute back

u = cos (c)

cos (c) = - \frac{2}{9}

cos (c) = - \frac{2}{9}

cos (c) = - \frac{2}{9} : c = arccos (- \frac{2}{9}) + 2 πn, c = - arccos (- \frac{2}{9}) + 2 πn

cos (c) = - \frac{2}{9}

Apply trig inverse properties

cos (c) = - \frac{2}{9}

General solutions for

cos (c) = - \frac{2}{9}

cos (x) = - a \Rightarrow x = arccos (- a) + 2 πn, x = - arccos (- a) + 2 πn

c = arccos (- \frac{2}{9}) + 2 πn, c = - arccos (- \frac{2}{9}) + 2 πn

c = arccos (- \frac{2}{9}) + 2 πn, c = - arccos (- \frac{2}{9}) + 2 πn

Combine all the solutions

c = arccos (- \frac{2}{9}) + 2 πn, c = - arccos (- \frac{2}{9}) + 2 πn

Show solutions in decimal form

c = 1.79488 \dots + 2 πn, c = - 1.79488 \dots + 2 πn

tan (x - \frac{π}{3}) = 1

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

1 = sin (2 t)

tan (2 x) + 2 sin (x) = 0

sin (\frac{x}{3}) = \frac{1}{2}, 0 \leq x \leq 2 π

What is the general solution for 10cos(c)+1=cos(c)-1 ?
The general solution for 10cos(c)+1=cos(c)-1 is c=1.79488…+2pin,c=-1.79488…+2pin