What is the general solution for 3cos(x)-1=4cos(x) ?

The general solution for 3cos(x)-1=4cos(x) is x=pi+2pin

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3cos(x)-1=4cos(x)

3 cos (x) - 1 = 4 cos (x)

x = π + 2 πn

Degrees

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

Solution steps

3 cos (x) - 1 = 4 cos (x)

Solve by substitution

3 cos (x) - 1 = 4 cos (x)

Let:

cos (x) = u

3 u - 1 = 4 u

3 u - 1 = 4 u : u = - 1

3 u - 1 = 4 u

Move

1

to the right side

3 u - 1 = 4 u

Add

1

to both sides

3 u - 1 + 1 = 4 u + 1

Simplify

3 u = 4 u + 1

3 u = 4 u + 1

Move

4 u

to the left side

3 u = 4 u + 1

Subtract

4 u

from both sides

3 u - 4 u = 4 u + 1 - 4 u

Simplify

- u = 1

- u = 1

Divide both sides by

- 1

- u = 1

Divide both sides by

- 1

\frac{- u}{- 1} = \frac{1}{- 1}

Simplify

u = - 1

u = - 1

Substitute back

u = cos (x)

cos (x) = - 1

cos (x) = - 1

cos (x) = - 1 : x = π + 2 πn

cos (x) = - 1

General solutions for

cos (x) = - 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}

x = π + 2 πn

x = π + 2 πn

Combine all the solutions

x = π + 2 πn

sin (x) = 666

- 2 cot (θ) = cot^{2} (θ) + 1

[sin (4 x) cos (x) - cos (4 x) sin (x)]^{2} = 1

cos (θ) = \frac{202}{331 \cdot 2 33}

5 sin (θ) + 12 cos (θ) = 13

What is the general solution for 3cos(x)-1=4cos(x) ?
The general solution for 3cos(x)-1=4cos(x) is x=pi+2pin