What is the general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 ?

The general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 is x=pin,x= pi/2+pin

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Popular >

2sin(x)cos(x)-sin(2x)cos(2x)=0

Solution

2 sin (x) cos (x) - sin (2 x) cos (2 x) = 0

Solution

x = πn, x = \frac{π}{2} + πn

Degrees

x = 0^{\circ} + 18 0^{\circ} n, x = 9 0^{\circ} + 18 0^{\circ} n

Solution steps

2 sin (x) cos (x) - sin (2 x) cos (2 x) = 0

Rewrite using trig identities

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

Use the Double Angle identity:

2 sin (x) cos (x) = sin (2 x)

= - cos (2 x) sin (2 x) + sin (2 x)

sin (2 x) - cos (2 x) sin (2 x) = 0

Factor

sin (2 x) - cos (2 x) sin (2 x) : - sin (2 x) (cos (2 x) - 1)

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

Factor out common term

- sin (2 x)

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

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

Solving each part separately

sin (2 x) = 0 or cos (2 x) - 1 = 0

sin (2 x) = 0 : x = πn, x = \frac{π}{2} + πn

sin (2 x) = 0

General solutions for

sin (2 x) = 0

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 = 0 + 2 πn, 2 x = π + 2 πn

2 x = 0 + 2 πn, 2 x = π + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

Solve

2 x = π + 2 πn : x = \frac{π}{2} + πn

2 x = π + 2 πn

Divide both sides by

2

2 x = π + 2 πn

Divide both sides by

2

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

Simplify

x = \frac{π}{2} + πn

x = \frac{π}{2} + πn

x = πn, x = \frac{π}{2} + πn

cos (2 x) - 1 = 0 : x = πn

cos (2 x) - 1 = 0

Move

1

to the right side

cos (2 x) - 1 = 0

Add

1

to both sides

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

Simplify

cos (2 x) = 1

cos (2 x) = 1

General solutions for

cos (2 x) = 1

cos (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} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{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} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

2 x = 0 + 2 πn

2 x = 0 + 2 πn

Solve

2 x = 0 + 2 πn : x = πn

2 x = 0 + 2 πn

0 + 2 πn = 2 πn

2 x = 2 πn

Divide both sides by

2

2 x = 2 πn

Divide both sides by

2

\frac{2 x}{2} = \frac{2 πn}{2}

Simplify

x = πn

x = πn

x = πn

Combine all the solutions

x = πn, x = \frac{π}{2} + πn

Popular Examples

cos(x)= 21/29

cos (x) = \frac{21}{29}

4/(sin(104.48))= 2/(sin(a))

\frac{4}{sin ( 104.4 8 ^{\circ} )} = \frac{2}{sin ( a )}

sin(x+(3pi)/2)-cos(x+pi/2)=1

sin (x + \frac{3 π}{2}) - cos (x + \frac{π}{2}) = 1

csc(x)-cot(x)=1

csc (x) - cot (x) = 1

sin(x)=-1/2 sin(2x)

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

Frequently Asked Questions (FAQ)

What is the general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 ?
The general solution for 2sin(x)cos(x)-sin(2x)cos(2x)=0 is x=pin,x= pi/2+pin