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

The general solution for sqrt(3)sin(2x)+cos(2x)=2cos(2x) is x= pi/(12)+(pin)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(3)sin(2x)+cos(2x)=2cos(2x)

Solution

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

Solution

x = \frac{π}{12} + \frac{πn}{2}

Degrees

x = 1 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Subtract

2 cos (2 x)

from both sides

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

Rewrite using trig identities

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

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

3 tan (2 x) - 1 = 0

3 tan (2 x) - 1 = 0

Move

1

to the right side

3 tan (2 x) - 1 = 0

Add

1

to both sides

3 tan (2 x) - 1 + 1 = 0 + 1

Simplify

3 tan (2 x) = 1

3 tan (2 x) = 1

Divide both sides by

3

3 tan (2 x) = 1

Divide both sides by

3

\frac{3 tan ( 2 x )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( 2 x )}{3} = \frac{1}{3}

Simplify

\frac{3 tan ( 2 x )}{3} : tan (2 x)

\frac{3 tan ( 2 x )}{3}

Cancel the common factor:

3

= tan (2 x)

Simplify

\frac{1}{3} : \frac{3}{3}

\frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

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

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= \frac{3}{3}

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

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

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

General solutions for

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

2 x = \frac{π}{6} + πn

2 x = \frac{π}{6} + πn

Solve

2 x = \frac{π}{6} + πn : x = \frac{π}{12} + \frac{πn}{2}

2 x = \frac{π}{6} + πn

Divide both sides by

2

2 x = \frac{π}{6} + πn

Divide both sides by

2

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

Simplify

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

Simplify

\frac{2 x}{2} : x

\frac{2 x}{2}

Divide the numbers:

\frac{2}{2} = 1

= x

Simplify

\frac{\frac{π}{6}}{2} + \frac{πn}{2} : \frac{π}{12} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} + \frac{πn}{2}

\frac{\frac{π}{6}}{2} = \frac{π}{12}

\frac{\frac{π}{6}}{2}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{π}{12}

= \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

x = \frac{π}{12} + \frac{πn}{2}

Popular Examples

sin(x)=-sin(x)cos(x)

sin (x) = - sin (x) cos (x)

tan^2(x/2)=1

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

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

cos (x) + sin^{2} (x) - cos^{2} (x) = 0

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

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

sin(x+pi/2)=(sqrt(3))/2

sin (x + \frac{π}{2}) = \frac{3}{2}

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sin(2x)+cos(2x)=2cos(2x) ?
The general solution for sqrt(3)sin(2x)+cos(2x)=2cos(2x) is x= pi/(12)+(pin)/2