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

The general solution for 2sin(2x)=1,interval(0,360) is No Solution for x\in\mathbb{R}

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2sin(2x)=1,interval(0,360)

Solution

2 sin (2 x) = 1, in t er v a l (0, 36 0^{\circ})

Solution

No Solution for x \in R

Solution steps

2 sin (2 x) = 1, in t er v a l (0, 36 0^{\circ})

Divide both sides by

2

2 sin (2 x) = 1

Divide both sides by

2

\frac{2 sin ( 2 x )}{2} = \frac{1}{2}

Simplify

sin (2 x) = \frac{1}{2}

sin (2 x) = \frac{1}{2}

General solutions for

sin (2 x) = \frac{1}{2}

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 = 3 0^{\circ} + 36 0^{\circ} n, 2 x = 15 0^{\circ} + 36 0^{\circ} n

2 x = 3 0^{\circ} + 36 0^{\circ} n, 2 x = 15 0^{\circ} + 36 0^{\circ} n

Solve

2 x = 3 0^{\circ} + 36 0^{\circ} k : x = 1 5^{\circ} + 18 0^{\circ} k

2 x = 3 0^{\circ} + 36 0^{\circ} k

Divide both sides by

2

2 x = 3 0^{\circ} + 36 0^{\circ} k

Divide both sides by

2

\frac{2 x}{2} = \frac{3 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} = \frac{3 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{3 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2} : 1 5^{\circ} + 18 0^{\circ} k

\frac{3 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

\frac{3 0 ^{\circ}}{2} = 1 5^{\circ}

\frac{3 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{18 0 ^{\circ}}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= 1 5^{\circ}

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

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

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} k

= 1 5^{\circ} + 18 0^{\circ} k

x = 1 5^{\circ} + 18 0^{\circ} k

x = 1 5^{\circ} + 18 0^{\circ} k

x = 1 5^{\circ} + 18 0^{\circ} k

Solve

2 x = 15 0^{\circ} + 36 0^{\circ} k : x = 7 5^{\circ} + 18 0^{\circ} k

2 x = 15 0^{\circ} + 36 0^{\circ} k

Divide both sides by

2

2 x = 15 0^{\circ} + 36 0^{\circ} k

Divide both sides by

2

\frac{2 x}{2} = \frac{15 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} = \frac{15 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{15 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2} : 7 5^{\circ} + 18 0^{\circ} k

\frac{15 0 ^{\circ}}{2} + \frac{36 0 ^{\circ} k}{2}

\frac{15 0 ^{\circ}}{2} = 7 5^{\circ}

\frac{15 0 ^{\circ}}{2}

Apply the fraction rule:

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

= \frac{90 0 ^{\circ}}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= 7 5^{\circ}

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

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

Divide the numbers:

\frac{2}{2} = 1

= 18 0^{\circ} k

= 7 5^{\circ} + 18 0^{\circ} k

x = 7 5^{\circ} + 18 0^{\circ} k

x = 7 5^{\circ} + 18 0^{\circ} k

x = 7 5^{\circ} + 18 0^{\circ} k

x = 1 5^{\circ} + 18 0^{\circ} k, x = 7 5^{\circ} + 18 0^{\circ} k

Solutions for the range

in t er v a l (0, 36 0^{\circ})

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for 2sin(2x)=1,interval(0,360) ?
The general solution for 2sin(2x)=1,interval(0,360) is No Solution for x\in\mathbb{R}