What is the general solution for 112=72+40cos(18t) ?

The general solution for 112=72+40cos(18t) is t=(pin)/9

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112=72+40cos(18t)

112 = 72 + 40 cos (18 t)

t = \frac{πn}{9}

Degrees

t = 0^{\circ} + 2 0^{\circ} n

Solution steps

112 = 72 + 40 cos (18 t)

Switch sides

72 + 40 cos (18 t) = 112

Move

72

to the right side

72 + 40 cos (18 t) = 112

Subtract

72

from both sides

72 + 40 cos (18 t) - 72 = 112 - 72

Simplify

40 cos (18 t) = 40

40 cos (18 t) = 40

Divide both sides by

40

40 cos (18 t) = 40

Divide both sides by

40

\frac{40 cos ( 18 t )}{40} = \frac{40}{40}

Simplify

cos (18 t) = 1

cos (18 t) = 1

General solutions for

cos (18 t) = 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}

18 t = 0 + 2 πn

18 t = 0 + 2 πn

Solve

18 t = 0 + 2 πn : t = \frac{πn}{9}

18 t = 0 + 2 πn

0 + 2 πn = 2 πn

18 t = 2 πn

Divide both sides by

18

18 t = 2 πn

Divide both sides by

18

\frac{18 t}{18} = \frac{2 πn}{18}

Simplify

t = \frac{πn}{9}

t = \frac{πn}{9}

t = \frac{πn}{9}

1 + sin (x) = (2) (cos (x))

cos (2 x) = - 0.32

sin (\frac{3 π}{2} - 0) = - cos (x)

sec (x) = - \frac{7}{4}

28 cos^{2} (θ) = 7

What is the general solution for 112=72+40cos(18t) ?
The general solution for 112=72+40cos(18t) is t=(pin)/9