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

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

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

Solution

sin (x) \cdot cos (x) = sin (2 x)

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

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

Subtract

sin (2 x)

from both sides

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

Rewrite using trig identities

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

Use the Double Angle identity:

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

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

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

Simplify

- sin (2 x) + \frac{sin ( 2 x )}{2} : - \frac{sin ( 2 x )}{2}

- sin (2 x) + \frac{sin ( 2 x )}{2}

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

Add similar elements:

sin (2 x) - 2 sin (2 x) = - sin (2 x)

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

Apply the fraction rule:

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

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

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

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

Multiply both sides by

2

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

Multiply both sides by

2

2 (- \frac{sin ( 2 x )}{2}) = 0 \cdot 2

Simplify

- sin (2 x) = 0

- sin (2 x) = 0

Divide both sides by

- 1

- sin (2 x) = 0

Divide both sides by

- 1

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

Simplify

sin (2 x) = 0

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

Popular Examples

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

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

sin(θ)=(-1)/(sqrt(2)),0<= θ<= 4pi

sin (θ) = \frac{- 1}{2}, 0 \leq θ \leq 4 π

0=2cos(θ)

0 = 2 cos (θ)

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

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

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

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

Frequently Asked Questions (FAQ)

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