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

The general solution for -sqrt(3)sin(2x)=cos(2x) is x=(5pi}{12}+\frac{pin)/2

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Popular >

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

Solution

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

Solution

x = \frac{5 π}{12} + \frac{πn}{2}

Degrees

x = 7 5^{\circ} + 9 0^{\circ} n

Solution steps

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

- 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}

Apply the fraction rule:

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

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

Cancel the common factor:

3

= tan (2 x)

Simplify

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

\frac{1}{- 3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

- \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}

= - \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{5 π}{6} + πn

2 x = \frac{5 π}{6} + πn

Solve

2 x = \frac{5 π}{6} + πn : x = \frac{5 π}{12} + \frac{πn}{2}

2 x = \frac{5 π}{6} + πn

Divide both sides by

2

2 x = \frac{5 π}{6} + πn

Divide both sides by

2

\frac{2 x}{2} = \frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

Simplify

\frac{2 x}{2} = \frac{\frac{5 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{5 π}{6}}{2} + \frac{πn}{2} : \frac{5 π}{12} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} + \frac{πn}{2}

\frac{\frac{5 π}{6}}{2} = \frac{5 π}{12}

\frac{\frac{5 π}{6}}{2}

Apply the fraction rule:

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

= \frac{5 π}{6 \cdot 2}

Multiply the numbers:

6 \cdot 2 = 12

= \frac{5 π}{12}

= \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

x = \frac{5 π}{12} + \frac{πn}{2}

Popular Examples

solvefor x,-sin(x)-cos(x)=0 solve for

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

cos(2θ)=((sqrt(57))/(11))

cos (2 θ) = (\frac{57}{11})

sin(x)cos(x^2)=sin(x),0<= x<= 2pi

sin (x) cos (x^{2}) = sin (x), 0 \leq x \leq 2 π

solvefor θ,(cos(θ))/(sin(θ))=-pi/6 solve for

θ, \frac{cos ( θ )}{sin ( θ )} = - \frac{π}{6}

tan(θ)=(10)/(-5.32)

tan (θ) = \frac{10}{- 5.32}

Frequently Asked Questions (FAQ)

What is the general solution for -sqrt(3)sin(2x)=cos(2x) ?
The general solution for -sqrt(3)sin(2x)=cos(2x) is x=(5pi}{12}+\frac{pin)/2