What is the general solution for sqrt(3)sin(3x)+3cos(3x)=0 ?

The general solution for sqrt(3)sin(3x)+3cos(3x)=0 is x=(2pi)/9+(pin)/3

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Popular >

sqrt(3)sin(3x)+3cos(3x)=0

Solution

3 sin (3 x) + 3 cos (3 x) = 0

Solution

x = \frac{2 π}{9} + \frac{πn}{3}

Degrees

x = 4 0^{\circ} + 6 0^{\circ} n

Solution steps

3 sin (3 x) + 3 cos (3 x) = 0

Rewrite using trig identities

3 sin (3 x) + 3 cos (3 x) = 0

Divide both sides by

cos (3 x), cos (3 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

3 tan (3 x) + 3 = 0

3 tan (3 x) + 3 = 0

Move

3

to the right side

3 tan (3 x) + 3 = 0

Subtract

3

from both sides

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

Simplify

3 tan (3 x) = - 3

3 tan (3 x) = - 3

Divide both sides by

3

3 tan (3 x) = - 3

Divide both sides by

3

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

Simplify

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

Simplify

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

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

Cancel the common factor:

3

= tan (3 x)

Simplify

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

\frac{- 3}{3}

Apply the fraction rule:

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

= - \frac{3}{3}

Apply radical rule:

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

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

Apply exponent rule:

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

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

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

Subtract the numbers:

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

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

Apply radical rule:

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

= - 3

tan (3 x) = - 3

tan (3 x) = - 3

tan (3 x) = - 3

General solutions for

tan (3 x) = - 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}

3 x = \frac{2 π}{3} + πn

3 x = \frac{2 π}{3} + πn

Solve

3 x = \frac{2 π}{3} + πn : x = \frac{2 π}{9} + \frac{πn}{3}

3 x = \frac{2 π}{3} + πn

Divide both sides by

3

3 x = \frac{2 π}{3} + πn

Divide both sides by

3

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

Simplify

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

Simplify

\frac{3 x}{3} : x

\frac{3 x}{3}

Divide the numbers:

\frac{3}{3} = 1

= x

Simplify

\frac{\frac{2 π}{3}}{3} + \frac{πn}{3} : \frac{2 π}{9} + \frac{πn}{3}

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

\frac{\frac{2 π}{3}}{3} = \frac{2 π}{9}

\frac{\frac{2 π}{3}}{3}

Apply the fraction rule:

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

= \frac{2 π}{3 \cdot 3}

Multiply the numbers:

3 \cdot 3 = 9

= \frac{2 π}{9}

= \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

x = \frac{2 π}{9} + \frac{πn}{3}

Popular Examples

tan(θ)=-4/5 , pi/2 <θ<pi

tan (θ) = - \frac{4}{5}, \frac{π}{2} < θ < π

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

cos (2 x) = sin (x), 0 \leq x < 2 π

solvefor θ,x=sec(θ)solve for

θ, x = sec (θ)

sin(x)+cos(x)=-1,0<= x<= 2pi

sin (x) + cos (x) = - 1, 0 \leq x \leq 2 π

229cos(a)+14715sin(a)=3959.55

229 cos (a) + 14715 sin (a) = 3959.55

Frequently Asked Questions (FAQ)

What is the general solution for sqrt(3)sin(3x)+3cos(3x)=0 ?
The general solution for sqrt(3)sin(3x)+3cos(3x)=0 is x=(2pi)/9+(pin)/3