What is the general solution for 8cos(12x)+4=-4 ?

The general solution for 8cos(12x)+4=-4 is x= pi/(12)+(pin)/6

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8cos(12x)+4=-4

8 cos (12 x) + 4 = - 4

x = \frac{π}{12} + \frac{πn}{6}

Degrees

x = 1 5^{\circ} + 3 0^{\circ} n

Solution steps

8 cos (12 x) + 4 = - 4

Move

4

to the right side

8 cos (12 x) + 4 = - 4

Subtract

4

from both sides

8 cos (12 x) + 4 - 4 = - 4 - 4

Simplify

8 cos (12 x) = - 8

8 cos (12 x) = - 8

Divide both sides by

8

8 cos (12 x) = - 8

Divide both sides by

8

\frac{8 cos ( 12 x )}{8} = \frac{- 8}{8}

Simplify

cos (12 x) = - 1

cos (12 x) = - 1

General solutions for

cos (12 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}

12 x = π + 2 πn

12 x = π + 2 πn

Solve

12 x = π + 2 πn : x = \frac{π}{12} + \frac{πn}{6}

12 x = π + 2 πn

Divide both sides by

12

12 x = π + 2 πn

Divide both sides by

12

\frac{12 x}{12} = \frac{π}{12} + \frac{2 πn}{12}

Simplify

x = \frac{π}{12} + \frac{πn}{6}

x = \frac{π}{12} + \frac{πn}{6}

x = \frac{π}{12} + \frac{πn}{6}

12 tan^{3} (x) = 4 tan (x)

cos (2 θ) = 3 sin (θ) - 1

sin (4 x) = - 1

cos (2 x) = sin (2 x)

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

What is the general solution for 8cos(12x)+4=-4 ?
The general solution for 8cos(12x)+4=-4 is x= pi/(12)+(pin)/6