What is the general solution for (sin(2x)+1)=0,0<= x<360 ?

The general solution for (sin(2x)+1)=0,0<= x<360 is x=135,x=315

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(sin(2x)+1)=0,0<= x<360

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

Radians

x = \frac{3 π}{4}, x = \frac{7 π}{4}

Solution steps

(sin (2 x) + 1) = 0, 0^{\circ} \leq x < 36 0^{\circ}

Move

1

to the right side

sin (2 x) + 1 = 0

Subtract

1

from both sides

sin (2 x) + 1 - 1 = 0 - 1

Simplify

sin (2 x) = - 1

sin (2 x) = - 1

General solutions for

sin (2 x) = - 1

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

2 x = 27 0^{\circ} + 36 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Solve

2 x = 27 0^{\circ} + 36 0^{\circ} n : x = 13 5^{\circ} + 18 0^{\circ} n

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

2 x = 27 0^{\circ} + 36 0^{\circ} n

Divide both sides by

2

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} = \frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2} : 13 5^{\circ} + 18 0^{\circ} n

\frac{27 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} n}{2}

\frac{27 0 ^{\circ}}{2} = 13 5^{\circ}

\frac{27 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{54 0 ^{\circ}}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= 13 5^{\circ}

\frac{36 0 ^{\circ} n}{2} = 18 0^{\circ} n

\frac{36 0 ^{\circ} n}{2}

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} n

= 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

x = 13 5^{\circ} + 18 0^{\circ} n

Solutions for the range

0 \leq x < 36 0^{\circ}

x = 13 5^{\circ}, x = 31 5^{\circ}

4 sin (x) + 2 = 2 sin (x)

y, - \frac{1}{2} cot (y) = \frac{t ^{2}}{2} + C

4 cos (2 t) + 1 = 3, 0, \frac{π}{2}

A, cos (\frac{π}{3} - A) = 2 sin (A)

- 3 + 3 = 26 cos (θ)

What is the general solution for (sin(2x)+1)=0,0<= x<360 ?
The general solution for (sin(2x)+1)=0,0<= x<360 is x=135,x=315