What is the general solution for 1+sin(5x)=0 ?

The general solution for 1+sin(5x)=0 is x=(3pi}{10}+\frac{2pin)/5

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1+sin(5x)=0

1 + sin (5 x) = 0

x = \frac{3 π}{10} + \frac{2 πn}{5}

Degrees

x = 5 4^{\circ} + 7 2^{\circ} n

Solution steps

1 + sin (5 x) = 0

Move

1

to the right side

1 + sin (5 x) = 0

Subtract

1

from both sides

1 + sin (5 x) - 1 = 0 - 1

Simplify

sin (5 x) = - 1

sin (5 x) = - 1

General solutions for

sin (5 x) = - 1

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}

5 x = \frac{3 π}{2} + 2 πn

5 x = \frac{3 π}{2} + 2 πn

Solve

5 x = \frac{3 π}{2} + 2 πn : x = \frac{3 π}{10} + \frac{2 πn}{5}

5 x = \frac{3 π}{2} + 2 πn

Divide both sides by

5

5 x = \frac{3 π}{2} + 2 πn

Divide both sides by

5

\frac{5 x}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 x}{5} = \frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

Simplify

\frac{5 x}{5} : x

\frac{5 x}{5}

Divide the numbers:

\frac{5}{5} = 1

= x

Simplify

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5} : \frac{3 π}{10} + \frac{2 πn}{5}

\frac{\frac{3 π}{2}}{5} + \frac{2 πn}{5}

\frac{\frac{3 π}{2}}{5} = \frac{3 π}{10}

\frac{\frac{3 π}{2}}{5}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

= \frac{3 π}{2 \cdot 5}

Multiply the numbers:

2 \cdot 5 = 10

= \frac{3 π}{10}

= \frac{3 π}{10} + \frac{2 πn}{5}

x = \frac{3 π}{10} + \frac{2 πn}{5}

x = \frac{3 π}{10} + \frac{2 πn}{5}

x = \frac{3 π}{10} + \frac{2 πn}{5}

x = \frac{3 π}{10} + \frac{2 πn}{5}

0 = \frac{π}{2} - 2 arctan (\frac{\frac{π}{2} - 2 - c}{\frac{π}{2} - c})

9 arcsin (x) = 3 π

2 sin (2 x + 10) - 3 = 0

cos (θ) = \frac{13}{6 \cdot 86}

\frac{1 + tan ( x )}{1 + cot ( x )} = 2

What is the general solution for 1+sin(5x)=0 ?
The general solution for 1+sin(5x)=0 is x=(3pi}{10}+\frac{2pin)/5