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

The general solution for 4cos(θ)=4sqrt(3)sin(θ) is θ= pi/6+pin

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4cos(θ)=4sqrt(3)sin(θ)

4 cos (θ) = 43 sin (θ)

θ = \frac{π}{6} + πn

Degrees

θ = 3 0^{\circ} + 18 0^{\circ} n

Solution steps

4 cos (θ) = 43 sin (θ)

Rewrite using trig identities

4 cos (θ) = 43 sin (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

4 = \frac{4 3 sin ( θ )}{cos ( θ )}

Use the basic trigonometric identity:

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

4 = 43 tan (θ)

Switch sides

43 tan (θ) = 4

43 tan (θ) = 4

Divide both sides by

43

43 tan (θ) = 4

Divide both sides by

43

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

Simplify

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

Simplify

\frac{4 3 tan ( θ )}{4 3} : tan (θ)

\frac{4 3 tan ( θ )}{4 3}

Divide the numbers:

\frac{4}{4} = 1

= \frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

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 (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

tan (θ) = \frac{3}{3}

General solutions for

tan (θ) = \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}

θ = \frac{π}{6} + πn

θ = \frac{π}{6} + πn

- 4 sin (θ) = 4 (sin (θ) - 1)

4 + 2 cos (θ) = 0

9.81 sin (θ) - 5.88 cos (θ) = 2.77

cos (θ) = π

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

What is the general solution for 4cos(θ)=4sqrt(3)sin(θ) ?
The general solution for 4cos(θ)=4sqrt(3)sin(θ) is θ= pi/6+pin