What is the general solution for sin(2θ)=3cos(2θ) ?

The general solution for sin(2θ)=3cos(2θ) is θ=(1.24904…)/2+(pin)/2

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sin(2θ)=3cos(2θ)

sin (2 θ) = 3 cos (2 θ)

θ = \frac{1.24904 \dots}{2} + \frac{πn}{2}

Degrees

θ = 35.78252 \dots^{\circ} + 9 0^{\circ} n

Solution steps

sin (2 θ) = 3 cos (2 θ)

Rewrite using trig identities

sin (2 θ) = 3 cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

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

Simplify

\frac{sin ( 2 θ )}{cos ( 2 θ )} = 3

Use the basic trigonometric identity:

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

tan (2 θ) = 3

tan (2 θ) = 3

Apply trig inverse properties

tan (2 θ) = 3

General solutions for

tan (2 θ) = 3

tan (x) = a \Rightarrow x = arctan (a) + πn

2 θ = arctan (3) + πn

2 θ = arctan (3) + πn

Solve

2 θ = arctan (3) + πn : θ = \frac{arctan ( 3 )}{2} + \frac{πn}{2}

2 θ = arctan (3) + πn

Divide both sides by

2

2 θ = arctan (3) + πn

Divide both sides by

2

\frac{2 θ}{2} = \frac{arctan ( 3 )}{2} + \frac{πn}{2}

Simplify

θ = \frac{arctan ( 3 )}{2} + \frac{πn}{2}

θ = \frac{arctan ( 3 )}{2} + \frac{πn}{2}

θ = \frac{arctan ( 3 )}{2} + \frac{πn}{2}

Show solutions in decimal form

θ = \frac{1.24904 \dots}{2} + \frac{πn}{2}

1 - cos (2 x) = sin (2 x)

cos (θ) = 5

2 cos^{2} (x) + 2 cos (x) = 0

x, tan (3 x) = 5 tan (x)

3 sin (2 t) = 0

What is the general solution for sin(2θ)=3cos(2θ) ?
The general solution for sin(2θ)=3cos(2θ) is θ=(1.24904…)/2+(pin)/2