What is the general solution for -3sin(2θ)=5cos(2θ),0<= 2θ<720 ?

The general solution for -3sin(2θ)=5cos(2θ),0<= 2θ<720 is No Solution for θ\in\mathbb{R}

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-3sin(2θ)=5cos(2θ),0<= 2θ<720

- 3 sin (2 θ) = 5 cos (2 θ), 0^{\circ} \leq 2 θ < 72 0^{\circ}

No Solution for θ \in R

Solution steps

- 3 sin (2 θ) = 5 cos (2 θ), 0^{\circ} \leq 2 θ < 72 0^{\circ}

Rewrite using trig identities

- 3 sin (2 θ) = 5 cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{- 3 sin ( 2 θ )}{cos ( 2 θ )} = \frac{5 cos ( 2 θ )}{cos ( 2 θ )}

Simplify

- \frac{3 sin ( 2 θ )}{cos ( 2 θ )} = 5

Use the basic trigonometric identity:

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

- 3 tan (2 θ) = 5

- 3 tan (2 θ) = 5

Divide both sides by

- 3

- 3 tan (2 θ) = 5

Divide both sides by

- 3

\frac{- 3 tan ( 2 θ )}{- 3} = \frac{5}{- 3}

Simplify

tan (2 θ) = - \frac{5}{3}

tan (2 θ) = - \frac{5}{3}

Apply trig inverse properties

tan (2 θ) = - \frac{5}{3}

General solutions for

tan (2 θ) = - \frac{5}{3}

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

2 θ = arctan (- \frac{5}{3}) + 18 0^{\circ} n

2 θ = arctan (- \frac{5}{3}) + 18 0^{\circ} n

Solve

2 θ = arctan (- \frac{5}{3}) + 18 0^{\circ} n : θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{18 0 ^{\circ} n}{2}

2 θ = arctan (- \frac{5}{3}) + 18 0^{\circ} n

Simplify

arctan (- \frac{5}{3}) + 18 0^{\circ} n : - arctan (\frac{5}{3}) + 18 0^{\circ} n

arctan (- \frac{5}{3}) + 18 0^{\circ} n

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{5}{3}) = - arctan (\frac{5}{3})

= - arctan (\frac{5}{3}) + 18 0^{\circ} n

2 θ = - arctan (\frac{5}{3}) + 18 0^{\circ} n

Divide both sides by

2

2 θ = - arctan (\frac{5}{3}) + 18 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{18 0 ^{\circ} n}{2}

Solutions for the range

0 \leq 2 θ < 72 0^{\circ}

No Solution for θ \in R

sin (4 x) = 1

2 sin^{2} (x) = 1 - cos (x)

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

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

3 sin (x) = 1 + cos (2 x)

What is the general solution for -3sin(2θ)=5cos(2θ),0<= 2θ<720 ?
The general solution for -3sin(2θ)=5cos(2θ),0<= 2θ<720 is No Solution for θ\in\mathbb{R}