What is the general solution for 4sqrt(3)sin(3θ+20)=4cos(3θ+20) ?

The general solution for 4sqrt(3)sin(3θ+20)=4cos(3θ+20) is θ=-20/3+pi/(18)+(pin)/3

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4sqrt(3)sin(3θ+20)=4cos(3θ+20)

Solution

43 sin (3 θ + 20) = 4 cos (3 θ + 20)

Solution

θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

Degrees

θ = - 371.97186 \dots^{\circ} + 6 0^{\circ} n

Solution steps

43 sin (3 θ + 20) = 4 cos (3 θ + 20)

Rewrite using trig identities

43 sin (3 θ + 20) = 4 cos (3 θ + 20)

Divide both sides by

cos (3 θ + 20), cos (3 θ + 20) \neq = 0

\frac{4 3 sin ( 3 θ + 20 )}{cos ( 3 θ + 20 )} = \frac{4 cos ( 3 θ + 20 )}{cos ( 3 θ + 20 )}

Simplify

\frac{4 3 sin ( 3 θ + 20 )}{cos ( 3 θ + 20 )} = 4

Use the basic trigonometric identity:

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

43 tan (3 θ + 20) = 4

43 tan (3 θ + 20) = 4

Divide both sides by

43

43 tan (3 θ + 20) = 4

Divide both sides by

43

\frac{4 3 tan ( 3 θ + 20 )}{4 3} = \frac{4}{4 3}

Simplify

\frac{4 3 tan ( 3 θ + 20 )}{4 3} = \frac{4}{4 3}

Simplify

\frac{4 3 tan ( 3 θ + 20 )}{4 3} : tan (3 θ + 20)

\frac{4 3 tan ( 3 θ + 20 )}{4 3}

Divide the numbers:

\frac{4}{4} = 1

= \frac{3 tan ( 3 θ + 20 )}{3}

Cancel the common factor:

3

= tan (3 θ + 20)

Simplify

\frac{4}{4 3} : \frac{3}{3}

\frac{4}{4 3}

Divide the numbers:

\frac{4}{4} = 1

= \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 (3 θ + 20) = \frac{3}{3}

tan (3 θ + 20) = \frac{3}{3}

tan (3 θ + 20) = \frac{3}{3}

General solutions for

tan (3 θ + 20) = \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}

3 θ + 20 = \frac{π}{6} + πn

3 θ + 20 = \frac{π}{6} + πn

Solve

3 θ + 20 = \frac{π}{6} + πn : θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

3 θ + 20 = \frac{π}{6} + πn

Move

20

to the right side

3 θ + 20 = \frac{π}{6} + πn

Subtract

20

from both sides

3 θ + 20 - 20 = \frac{π}{6} + πn - 20

Simplify

3 θ = \frac{π}{6} + πn - 20

3 θ = \frac{π}{6} + πn - 20

Divide both sides by

3

3 θ = \frac{π}{6} + πn - 20

Divide both sides by

3

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{πn}{3} - \frac{20}{3}

Simplify

\frac{3 θ}{3} = \frac{\frac{π}{6}}{3} + \frac{πn}{3} - \frac{20}{3}

Simplify

\frac{3 θ}{3} : θ

\frac{3 θ}{3}

Divide the numbers:

\frac{3}{3} = 1

= θ

Simplify

\frac{\frac{π}{6}}{3} + \frac{πn}{3} - \frac{20}{3} : - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

\frac{\frac{π}{6}}{3} + \frac{πn}{3} - \frac{20}{3}

Group like terms

= - \frac{20}{3} + \frac{πn}{3} + \frac{\frac{π}{6}}{3}

\frac{\frac{π}{6}}{3} = \frac{π}{18}

\frac{\frac{π}{6}}{3}

Apply the fraction rule:

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

= \frac{π}{6 \cdot 3}

Multiply the numbers:

6 \cdot 3 = 18

= \frac{π}{18}

= - \frac{20}{3} + \frac{πn}{3} + \frac{π}{18}

Group like terms

= - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

θ = - \frac{20}{3} + \frac{π}{18} + \frac{πn}{3}

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Frequently Asked Questions (FAQ)

What is the general solution for 4sqrt(3)sin(3θ+20)=4cos(3θ+20) ?
The general solution for 4sqrt(3)sin(3θ+20)=4cos(3θ+20) is θ=-20/3+pi/(18)+(pin)/3