What is the general solution for -2cos(b)+4=cos(b)+6 ?

The general solution for -2cos(b)+4=cos(b)+6 is b=2.30052…+2pin,b=-2.30052…+2pin

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-2cos(b)+4=cos(b)+6

- 2 cos (b) + 4 = cos (b) + 6

b = 2.30052 \dots + 2 πn, b = - 2.30052 \dots + 2 πn

Degrees

b = 131.81031 \dots^{\circ} + 36 0^{\circ} n, b = - 131.81031 \dots^{\circ} + 36 0^{\circ} n

Solution steps

- 2 cos (b) + 4 = cos (b) + 6

Solve by substitution

- 2 cos (b) + 4 = cos (b) + 6

Let:

cos (b) = u

- 2 u + 4 = u + 6

- 2 u + 4 = u + 6 : u = - \frac{2}{3}

- 2 u + 4 = u + 6

Move

4

to the right side

- 2 u + 4 = u + 6

Subtract

4

from both sides

- 2 u + 4 - 4 = u + 6 - 4

Simplify

- 2 u = u + 2

- 2 u = u + 2

Move

u

to the left side

- 2 u = u + 2

Subtract

u

from both sides

- 2 u - u = u + 2 - u

Simplify

- 3 u = 2

- 3 u = 2

Divide both sides by

- 3

- 3 u = 2

Divide both sides by

- 3

\frac{- 3 u}{- 3} = \frac{2}{- 3}

Simplify

u = - \frac{2}{3}

u = - \frac{2}{3}

Substitute back

u = cos (b)

cos (b) = - \frac{2}{3}

cos (b) = - \frac{2}{3}

cos (b) = - \frac{2}{3} : b = arccos (- \frac{2}{3}) + 2 πn, b = - arccos (- \frac{2}{3}) + 2 πn

cos (b) = - \frac{2}{3}

Apply trig inverse properties

cos (b) = - \frac{2}{3}

General solutions for

cos (b) = - \frac{2}{3}

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

b = arccos (- \frac{2}{3}) + 2 πn, b = - arccos (- \frac{2}{3}) + 2 πn

b = arccos (- \frac{2}{3}) + 2 πn, b = - arccos (- \frac{2}{3}) + 2 πn

Combine all the solutions

b = arccos (- \frac{2}{3}) + 2 πn, b = - arccos (- \frac{2}{3}) + 2 πn

Show solutions in decimal form

b = 2.30052 \dots + 2 πn, b = - 2.30052 \dots + 2 πn

25 csc^{2} (θ) - 4 = 0

tan (θ) = \frac{1}{5}

3 cos (2 x) = 2

3 cos (2 x) = 0

3 cos (x) - 3 sin (x) = 0

What is the general solution for -2cos(b)+4=cos(b)+6 ?
The general solution for -2cos(b)+4=cos(b)+6 is b=2.30052…+2pin,b=-2.30052…+2pin