What is the general solution for 4cos(x)=3+8cos(x) ?

The general solution for 4cos(x)=3+8cos(x) is x=2.41885…+2pin,x=-2.41885…+2pin

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4cos(x)=3+8cos(x)

4 cos (x) = 3 + 8 cos (x)

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

Degrees

x = 138.59037 \dots^{\circ} + 36 0^{\circ} n, x = - 138.59037 \dots^{\circ} + 36 0^{\circ} n

Solution steps

4 cos (x) = 3 + 8 cos (x)

Solve by substitution

4 cos (x) = 3 + 8 cos (x)

Let:

cos (x) = u

4 u = 3 + 8 u

4 u = 3 + 8 u : u = - \frac{3}{4}

4 u = 3 + 8 u

Move

8 u

to the left side

4 u = 3 + 8 u

Subtract

8 u

from both sides

4 u - 8 u = 3 + 8 u - 8 u

Simplify

- 4 u = 3

- 4 u = 3

Divide both sides by

- 4

- 4 u = 3

Divide both sides by

- 4

\frac{- 4 u}{- 4} = \frac{3}{- 4}

Simplify

u = - \frac{3}{4}

u = - \frac{3}{4}

Substitute back

u = cos (x)

cos (x) = - \frac{3}{4}

cos (x) = - \frac{3}{4}

cos (x) = - \frac{3}{4} : x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

cos (x) = - \frac{3}{4}

Apply trig inverse properties

cos (x) = - \frac{3}{4}

General solutions for

cos (x) = - \frac{3}{4}

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

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Combine all the solutions

x = arccos (- \frac{3}{4}) + 2 πn, x = - arccos (- \frac{3}{4}) + 2 πn

Show solutions in decimal form

x = 2.41885 \dots + 2 πn, x = - 2.41885 \dots + 2 πn

sin^{2} (θ) - 3 sin (θ) = 0

cos^{2} (x) = \frac{1}{5}

- 3 cos (2 θ) + 5 = 7 sin (θ) + 5

3 + 3 cos (x) = 2 sin^{2} (x)

cos (θ) = - sin (θ)

What is the general solution for 4cos(x)=3+8cos(x) ?
The general solution for 4cos(x)=3+8cos(x) is x=2.41885…+2pin,x=-2.41885…+2pin