What is the general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 ?

The general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 is x= pi/6+(pin)/4

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sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2

Solution

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) = \frac{1}{2}

Solution

x = \frac{π}{6} + \frac{πn}{4}

Degrees

x = 3 0^{\circ} + 4 5^{\circ} n

Solution steps

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) = \frac{1}{2}

Subtract

\frac{1}{2}

from both sides

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) - \frac{1}{2} = 0

Simplify

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) - \frac{1}{2} : \frac{2 3 sin ^{2} ( 2 x ) - sin ( 4 x ) - 3}{2 3}

sin^{2} (2 x) - \frac{1}{2 3} sin (4 x) - \frac{1}{2}

\frac{1}{2 3} sin (4 x) = \frac{sin ( 4 x )}{2 3}

\frac{1}{2 3} sin (4 x)

Multiply fractions:

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

= \frac{1 \cdot sin ( 4 x )}{2 3}

Multiply:

1 \cdot sin (4 x) = sin (4 x)

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

= sin^{2} (2 x) - \frac{sin ( 4 x )}{2 3} - \frac{1}{2}

Convert element to fraction:

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

= \frac{sin ^{2} ( 2 x )}{1} - \frac{sin ( 4 x )}{2 3} - \frac{1}{2}

Least Common Multiplier of

1, 23, 2 : 23

1, 23, 2

Lowest Common Multiplier (LCM)

Least Common Multiplier of

1, 2, 2 : 2

1, 2, 2

Least Common Multiplier (LCM)

Prime factorization of

1

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Prime factorization of

2 : 2

2

2

is a prime number, therefore no factorization is possible

= 2

Compute a number comprised of factors that appear in at least one of the following:

1, 2, 2

= 2

Multiply the numbers:

2 = 2

= 2

Compute an expression comprised of factors that appear in at least one of the factored expressions

= 23

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

23

For

\frac{sin ^{2} ( 2 x )}{1} :

multiply the denominator and numerator by

23

\frac{sin ^{2} ( 2 x )}{1} = \frac{sin ^{2} ( 2 x ) \cdot 2 3}{1 \cdot 2 3} = \frac{sin ^{2} ( 2 x ) \cdot 2 3}{2 3}

For

\frac{1}{2} :

multiply the denominator and numerator by

3

\frac{1}{2} = \frac{1 \cdot 3}{2 3} = \frac{3}{2 3}

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

Since the denominators are equal, combine the fractions:

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

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

\frac{2 3 sin ^{2} ( 2 x ) - sin ( 4 x ) - 3}{2 3} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

23 sin^{2} (2 x) - sin (4 x) - 3 = 0

Rewrite using trig identities

- sin (4 x) - 3 + 2 sin^{2} (2 x) 3

Use the Double Angle identity:

1 - 2 sin^{2} (x) = cos (2 x)

- 2 sin^{2} (x) = cos (2 x) - 1

= - sin (4 x) - 3 - (cos (2 \cdot 2 x) - 1) 3

Simplify

- sin (4 x) - 3 - (cos (2 \cdot 2 x) - 1) 3 : - sin (4 x) - 3 cos (4 x)

- sin (4 x) - 3 - (cos (2 \cdot 2 x) - 1) 3

Multiply the numbers:

2 \cdot 2 = 4

= - sin (4 x) - 3 - 3 (cos (4 x) - 1)

Expand

- 3 (cos (4 x) - 1) : - 3 cos (4 x) + 3

- 3 (cos (4 x) - 1)

Apply the distributive law:

a (b - c) = ab - a c

a = - 3, b = cos (4 x), c = 1

= - 3 cos (4 x) - (- 3) \cdot 1

Apply minus-plus rules

- (- a) = a

= - 3 cos (4 x) + 1 \cdot 3

Multiply:

1 \cdot 3 = 3

= - 3 cos (4 x) + 3

= - sin (4 x) - 3 - 3 cos (4 x) + 3

Simplify

- sin (4 x) - 3 - 3 cos (4 x) + 3 : - sin (4 x) - 3 cos (4 x)

- sin (4 x) - 3 - 3 cos (4 x) + 3

Group like terms

= - sin (4 x) - 3 cos (4 x) - 3 + 3

Add similar elements:

- 3 + 3 = 0

= - sin (4 x) - 3 cos (4 x)

= - sin (4 x) - 3 cos (4 x)

= - sin (4 x) - 3 cos (4 x)

Divide both sides by

cos (4 x), cos (4 x) \neq = 0

\frac{- sin ( 4 x ) - cos ( 4 x ) 3}{cos ( 4 x )} = \frac{0}{cos ( 4 x )}

Simplify

- \frac{sin ( 4 x )}{cos ( 4 x )} - 3 = 0

Use the basic trigonometric identity:

\frac{sin ( x )}{cos ( x )} = tan (x)

- tan (4 x) - 3 = 0

- tan (4 x) - 3 = 0

Move

3

to the right side

- tan (4 x) - 3 = 0

Add

3

to both sides

- tan (4 x) - 3 + 3 = 0 + 3

Simplify

- tan (4 x) = 3

- tan (4 x) = 3

Divide both sides by

- 1

- tan (4 x) = 3

Divide both sides by

- 1

\frac{- tan ( 4 x )}{- 1} = \frac{3}{- 1}

Simplify

tan (4 x) = - 3

tan (4 x) = - 3

General solutions for

tan (4 x) = - 3

tan (x)

periodicity table with

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} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

4 x = \frac{2 π}{3} + πn

4 x = \frac{2 π}{3} + πn

Solve

4 x = \frac{2 π}{3} + πn : x = \frac{π}{6} + \frac{πn}{4}

4 x = \frac{2 π}{3} + πn

Divide both sides by

4

4 x = \frac{2 π}{3} + πn

Divide both sides by

4

\frac{4 x}{4} = \frac{\frac{2 π}{3}}{4} + \frac{πn}{4}

Simplify

\frac{4 x}{4} = \frac{\frac{2 π}{3}}{4} + \frac{πn}{4}

Simplify

\frac{4 x}{4} : x

\frac{4 x}{4}

Divide the numbers:

\frac{4}{4} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{4} + \frac{πn}{4} : \frac{π}{6} + \frac{πn}{4}

\frac{\frac{2 π}{3}}{4} + \frac{πn}{4}

\frac{\frac{2 π}{3}}{4} = \frac{π}{6}

\frac{\frac{2 π}{3}}{4}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 4}

Multiply the numbers:

3 \cdot 4 = 12

= \frac{2 π}{12}

Cancel the common factor:

2

= \frac{π}{6}

= \frac{π}{6} + \frac{πn}{4}

x = \frac{π}{6} + \frac{πn}{4}

x = \frac{π}{6} + \frac{πn}{4}

x = \frac{π}{6} + \frac{πn}{4}

x = \frac{π}{6} + \frac{πn}{4}

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Frequently Asked Questions (FAQ)

What is the general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 ?
The general solution for sin^2(2x)-1/(2sqrt(3))sin(4x)= 1/2 is x= pi/6+(pin)/4