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

The general solution for sin(x)cos(2x)-cos(x)sin(2x)=(sqrt(3))/2 is x=-pi/3-2pin,x=-(2pi)/3-2pin

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sin(x)cos(2x)-cos(x)sin(2x)=(sqrt(3))/2

Solution

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

Solution

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

Degrees

x = - 6 0^{\circ} - 36 0^{\circ} n, x = - 12 0^{\circ} - 36 0^{\circ} n

Solution steps

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

Rewrite using trig identities

sin (x) cos (2 x) - cos (x) sin (2 x)

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (x - 2 x)

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

General solutions for

sin (x - 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}

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

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

Solve

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

x - 2 x = \frac{π}{3} + 2 πn

Add similar elements:

x - 2 x = - x

- x = \frac{π}{3} + 2 πn

Divide both sides by

- 1

- x = \frac{π}{3} + 2 πn

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{\frac{π}{3}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- x}{- 1} = \frac{\frac{π}{3}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{\frac{π}{3}}{- 1} + \frac{2 πn}{- 1} : - \frac{π}{3} - 2 πn

\frac{\frac{π}{3}}{- 1} + \frac{2 πn}{- 1}

\frac{\frac{π}{3}}{- 1} = - \frac{π}{3}

\frac{\frac{π}{3}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{π}{3}}{1}

Apply the fraction rule:

\frac{a}{1} = a

\frac{\frac{π}{3}}{1} = \frac{π}{3}

= - \frac{π}{3}

= - \frac{π}{3} + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - \frac{π}{3} - 2 πn

x = - \frac{π}{3} - 2 πn

x = - \frac{π}{3} - 2 πn

x = - \frac{π}{3} - 2 πn

Solve

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

x - 2 x = \frac{2 π}{3} + 2 πn

Add similar elements:

x - 2 x = - x

- x = \frac{2 π}{3} + 2 πn

Divide both sides by

- 1

- x = \frac{2 π}{3} + 2 πn

Divide both sides by

- 1

\frac{- x}{- 1} = \frac{\frac{2 π}{3}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- x}{- 1} = \frac{\frac{2 π}{3}}{- 1} + \frac{2 πn}{- 1}

Simplify

\frac{- x}{- 1} : x

\frac{- x}{- 1}

Apply the fraction rule:

\frac{- a}{- b} = \frac{a}{b}

= \frac{x}{1}

Apply rule

\frac{a}{1} = a

= x

Simplify

\frac{\frac{2 π}{3}}{- 1} + \frac{2 πn}{- 1} : - \frac{2 π}{3} - 2 πn

\frac{\frac{2 π}{3}}{- 1} + \frac{2 πn}{- 1}

\frac{\frac{2 π}{3}}{- 1} = - \frac{2 π}{3}

\frac{\frac{2 π}{3}}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{\frac{2 π}{3}}{1}

Apply the fraction rule:

\frac{a}{1} = a

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

= - \frac{2 π}{3}

= - \frac{2 π}{3} + \frac{2 πn}{- 1}

\frac{2 πn}{- 1} = - 2 πn

\frac{2 πn}{- 1}

Apply the fraction rule:

\frac{a}{- b} = - \frac{a}{b}

= - \frac{2 πn}{1}

Apply rule

\frac{a}{1} = a

= - 2 πn

= - \frac{2 π}{3} - 2 πn

x = - \frac{2 π}{3} - 2 πn

x = - \frac{2 π}{3} - 2 πn

x = - \frac{2 π}{3} - 2 πn

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

Graph

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Frequently Asked Questions (FAQ)

What is the general solution for sin(x)cos(2x)-cos(x)sin(2x)=(sqrt(3))/2 ?
The general solution for sin(x)cos(2x)-cos(x)sin(2x)=(sqrt(3))/2 is x=-pi/3-2pin,x=-(2pi)/3-2pin