What is the general solution for 6sin(20x)+1=1 ?

The general solution for 6sin(20x)+1=1 is x=(pin)/(10),x= pi/(20)+(pin)/(10)

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6sin(20x)+1=1

6 sin (20 x) + 1 = 1

x = \frac{πn}{10}, x = \frac{π}{20} + \frac{πn}{10}

Degrees

x = 0^{\circ} + 1 8^{\circ} n, x = 9^{\circ} + 1 8^{\circ} n

Solution steps

6 sin (20 x) + 1 = 1

Subtract

1

from both sides

6 sin (20 x) + 1 - 1 = 1 - 1

Simplify

6 sin (20 x) = 0

Divide both sides by

6

6 sin (20 x) = 0

Divide both sides by

6

\frac{6 sin ( 20 x )}{6} = \frac{0}{6}

Simplify

sin (20 x) = 0

sin (20 x) = 0

General solutions for

sin (20 x) = 0

sin (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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{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} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

20 x = 0 + 2 πn, 20 x = π + 2 πn

20 x = 0 + 2 πn, 20 x = π + 2 πn

Solve

20 x = 0 + 2 πn : x = \frac{πn}{10}

20 x = 0 + 2 πn

0 + 2 πn = 2 πn

20 x = 2 πn

Divide both sides by

20

20 x = 2 πn

Divide both sides by

20

\frac{20 x}{20} = \frac{2 πn}{20}

Simplify

x = \frac{πn}{10}

x = \frac{πn}{10}

Solve

20 x = π + 2 πn : x = \frac{π}{20} + \frac{πn}{10}

20 x = π + 2 πn

Divide both sides by

20

20 x = π + 2 πn

Divide both sides by

20

\frac{20 x}{20} = \frac{π}{20} + \frac{2 πn}{20}

Simplify

x = \frac{π}{20} + \frac{πn}{10}

x = \frac{π}{20} + \frac{πn}{10}

x = \frac{πn}{10}, x = \frac{π}{20} + \frac{πn}{10}

sin (x) = 1 - cos (x)

tan (t) = 3

2 cos^{2} (x) - 9 cos (x) - 5 = 0

2 sec^{2} (x) = 3 - tan (x)

2 sin^{2} (θ) - sin (θ) = 1

What is the general solution for 6sin(20x)+1=1 ?
The general solution for 6sin(20x)+1=1 is x=(pin)/(10),x= pi/(20)+(pin)/(10)