What is the general solution for 4sin(2x)-sqrt(12)=0,0<= x<= 2pi ?

Q: What is the general solution for 4sin(2x)-sqrt(12)=0,0<= x<= 2pi ?

The general solution for 4sin(2x)-sqrt(12)=0,0<= x<= 2pi is x= pi/6 ,x= pi/3 ,x=(7pi)/6 ,x=(4pi)/3

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4sin(2x)-sqrt(12)=0,0<= x<= 2pi

Solution

4 sin (2 x) - 12 = 0, 0 \leq x \leq 2 π

Solution

x = \frac{π}{6}, x = \frac{π}{3}, x = \frac{7 π}{6}, x = \frac{4 π}{3}

Degrees

x = 3 0^{\circ}, x = 6 0^{\circ}, x = 21 0^{\circ}, x = 24 0^{\circ}

Solution steps

4 sin (2 x) - 12 = 0, 0 \leq x \leq 2 π

12 = 23

12

Prime factorization of

12 : 2^{2} \cdot 3

12

12

divides by

212 = 6 \cdot 2

= 2 \cdot 6

6

divides by

26 = 3 \cdot 2

= 2 \cdot 2 \cdot 3

2, 3

are all prime numbers, therefore no further factorization is possible

= 2 \cdot 2 \cdot 3

= 2^{2} \cdot 3

= 2^{2} \cdot 3

Apply radical rule:

= 3 2^{2}

Apply radical rule:

2^{2} = 2

= 23

4 sin (2 x) - 23 = 0

Move

23

to the right side

4 sin (2 x) - 23 = 0

Add

23

to both sides

4 sin (2 x) - 23 + 23 = 0 + 23

Simplify

4 sin (2 x) = 23

4 sin (2 x) = 23

Divide both sides by

4

4 sin (2 x) = 23

Divide both sides by

4

\frac{4 sin ( 2 x )}{4} = \frac{2 3}{4}

Simplify

sin (2 x) = \frac{3}{2}

sin (2 x) = \frac{3}{2}

General solutions for

sin (2 x) = \frac{3}{2}

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}

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

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

Solve

2 x = \frac{π}{3} + 2 πn : x = \frac{π}{6} + πn

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{3}}{2} + \frac{2 πn}{2} : \frac{π}{6} + πn

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

\frac{\frac{π}{3}}{2} = \frac{π}{6}

\frac{\frac{π}{3}}{2}

Apply the fraction rule:

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

= \frac{π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{π}{6}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

x = \frac{π}{6} + πn

Solve

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

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

2 x = \frac{2 π}{3} + 2 πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

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

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

\frac{\frac{2 π}{3}}{2} = \frac{π}{3}

\frac{\frac{2 π}{3}}{2}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 2}

Multiply the numbers:

3 \cdot 2 = 6

= \frac{2 π}{6}

Cancel the common factor:

2

= \frac{π}{3}

\frac{2 πn}{2} = πn

\frac{2 πn}{2}

Divide the numbers:

\frac{2}{2} = 1

= πn

= \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{3} + πn

x = \frac{π}{6} + πn, x = \frac{π}{3} + πn

Solutions for the range

0 \leq x \leq 2 π

x = \frac{π}{6}, x = \frac{π}{3}, x = \frac{7 π}{6}, x = \frac{4 π}{3}

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sin (x) - 0.5 \cdot cos (x) = 0.768

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\frac{( 176 ) ^{2}}{300 \cdot 32.2} = tan (θ + 30.963 7^{\circ})

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Frequently Asked Questions (FAQ)

What is the general solution for 4sin(2x)-sqrt(12)=0,0<= x<= 2pi ?
The general solution for 4sin(2x)-sqrt(12)=0,0<= x<= 2pi is x= pi/6 ,x= pi/3 ,x=(7pi)/6 ,x=(4pi)/3