What is the general solution for 5sin(2θ)=7cos(2θ),0<= 2θ<720 ?

The general solution for 5sin(2θ)=7cos(2θ),0<= 2θ<720 is No Solution for θ\in\mathbb{R}

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5sin(2θ)=7cos(2θ),0<= 2θ<720

5 sin (2 θ) = 7 cos (2 θ), 0^{\circ} \leq 2 θ < 72 0^{\circ}

No Solution for θ \in R

Solution steps

5 sin (2 θ) = 7 cos (2 θ), 0^{\circ} \leq 2 θ < 72 0^{\circ}

Rewrite using trig identities

5 sin (2 θ) = 7 cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} = \frac{7 cos ( 2 θ )}{cos ( 2 θ )}

Simplify

\frac{5 sin ( 2 θ )}{cos ( 2 θ )} = 7

Use the basic trigonometric identity:

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

5 tan (2 θ) = 7

5 tan (2 θ) = 7

Divide both sides by

5

5 tan (2 θ) = 7

Divide both sides by

5

\frac{5 tan ( 2 θ )}{5} = \frac{7}{5}

Simplify

tan (2 θ) = \frac{7}{5}

tan (2 θ) = \frac{7}{5}

Apply trig inverse properties

tan (2 θ) = \frac{7}{5}

General solutions for

tan (2 θ) = \frac{7}{5}

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

2 θ = arctan (\frac{7}{5}) + 18 0^{\circ} n

2 θ = arctan (\frac{7}{5}) + 18 0^{\circ} n

Solve

2 θ = arctan (\frac{7}{5}) + 18 0^{\circ} n : θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

2 θ = arctan (\frac{7}{5}) + 18 0^{\circ} n

Divide both sides by

2

2 θ = arctan (\frac{7}{5}) + 18 0^{\circ} n

Divide both sides by

2

\frac{2 θ}{2} = \frac{arctan ( \frac{7}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Simplify

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

θ = \frac{arctan ( \frac{7}{5} )}{2} + \frac{18 0 ^{\circ} n}{2}

Solutions for the range

0 \leq 2 θ < 72 0^{\circ}

No Solution for θ \in R

- 7 sin^{2} (θ) + 4 sin (θ) + 7 = 0

- 2 sin^{2} (x) = - 3 sin (x) + 1

sin (x) = \frac{6}{10}

sin (3 x) = 3 cos (3 x)

(9 sin (2 x) + 9 cos (2 x))^{2} = 81

What is the general solution for 5sin(2θ)=7cos(2θ),0<= 2θ<720 ?
The general solution for 5sin(2θ)=7cos(2θ),0<= 2θ<720 is No Solution for θ\in\mathbb{R}