What is the general solution for 4sin(3θ)=7cos(3θ),0<= 3θ<720 ?

The general solution for 4sin(3θ)=7cos(3θ),0<= 3θ<720 is No Solution for θ\in\mathbb{R}

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4sin(3θ)=7cos(3θ),0<= 3θ<720

4 sin (3 θ) = 7 cos (3 θ), 0^{\circ} \leq 3 θ < 72 0^{\circ}

No Solution for θ \in R

Solution steps

4 sin (3 θ) = 7 cos (3 θ), 0^{\circ} \leq 3 θ < 72 0^{\circ}

Rewrite using trig identities

4 sin (3 θ) = 7 cos (3 θ)

Divide both sides by

cos (3 θ), cos (3 θ) \neq = 0

\frac{4 sin ( 3 θ )}{cos ( 3 θ )} = \frac{7 cos ( 3 θ )}{cos ( 3 θ )}

Simplify

\frac{4 sin ( 3 θ )}{cos ( 3 θ )} = 7

Use the basic trigonometric identity:

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

4 tan (3 θ) = 7

4 tan (3 θ) = 7

Divide both sides by

4

4 tan (3 θ) = 7

Divide both sides by

4

\frac{4 tan ( 3 θ )}{4} = \frac{7}{4}

Simplify

tan (3 θ) = \frac{7}{4}

tan (3 θ) = \frac{7}{4}

Apply trig inverse properties

tan (3 θ) = \frac{7}{4}

General solutions for

tan (3 θ) = \frac{7}{4}

tan (x) = a \Rightarrow x = arctan (a) + 18 0^{\circ} n

3 θ = arctan (\frac{7}{4}) + 18 0^{\circ} n

3 θ = arctan (\frac{7}{4}) + 18 0^{\circ} n

Solve

3 θ = arctan (\frac{7}{4}) + 18 0^{\circ} n : θ = \frac{arctan ( \frac{7}{4} )}{3} + \frac{18 0 ^{\circ} n}{3}

3 θ = arctan (\frac{7}{4}) + 18 0^{\circ} n

Divide both sides by

3

3 θ = arctan (\frac{7}{4}) + 18 0^{\circ} n

Divide both sides by

3

\frac{3 θ}{3} = \frac{arctan ( \frac{7}{4} )}{3} + \frac{18 0 ^{\circ} n}{3}

Simplify

θ = \frac{arctan ( \frac{7}{4} )}{3} + \frac{18 0 ^{\circ} n}{3}

θ = \frac{arctan ( \frac{7}{4} )}{3} + \frac{18 0 ^{\circ} n}{3}

θ = \frac{arctan ( \frac{7}{4} )}{3} + \frac{18 0 ^{\circ} n}{3}

Solutions for the range

0 \leq 3 θ < 72 0^{\circ}

No Solution for θ \in R

cot (θ) = 4

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

0 = - \frac{1}{7} cos (7 t)

cos (α) = \frac{5}{13}

tan (x) = 3, 0 \leq x \leq 2 π

What is the general solution for 4sin(3θ)=7cos(3θ),0<= 3θ<720 ?
The general solution for 4sin(3θ)=7cos(3θ),0<= 3θ<720 is No Solution for θ\in\mathbb{R}