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

The general solution for 3sin(2θ)=2cos(2θ) is θ=(0.58800…)/2+(pin)/2

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3sin(2θ)=2cos(2θ)

3 sin (2 θ) = 2 cos (2 θ)

θ = \frac{0.58800 \dots}{2} + \frac{πn}{2}

Degrees

θ = 16.84503 \dots^{\circ} + 9 0^{\circ} n

Solution steps

3 sin (2 θ) = 2 cos (2 θ)

Rewrite using trig identities

3 sin (2 θ) = 2 cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

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

Simplify

\frac{3 sin ( 2 θ )}{cos ( 2 θ )} = 2

Use the basic trigonometric identity:

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

3 tan (2 θ) = 2

3 tan (2 θ) = 2

Divide both sides by

3

3 tan (2 θ) = 2

Divide both sides by

3

\frac{3 tan ( 2 θ )}{3} = \frac{2}{3}

Simplify

tan (2 θ) = \frac{2}{3}

tan (2 θ) = \frac{2}{3}

Apply trig inverse properties

tan (2 θ) = \frac{2}{3}

General solutions for

tan (2 θ) = \frac{2}{3}

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

2 θ = arctan (\frac{2}{3}) + πn

2 θ = arctan (\frac{2}{3}) + πn

Solve

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

2 θ = arctan (\frac{2}{3}) + πn

Divide both sides by

2

2 θ = arctan (\frac{2}{3}) + πn

Divide both sides by

2

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

Simplify

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

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

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

Show solutions in decimal form

θ = \frac{0.58800 \dots}{2} + \frac{πn}{2}

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

cos (θ) = cos (1)

tan (\frac{x}{2}) = \pm \frac{1 - cos ( x )}{1 + cos ( x )}

sin^{2} (x) - cos^{2} (x) = 0.5

sec (x) - tan (x) = 1

What is the general solution for 3sin(2θ)=2cos(2θ) ?
The general solution for 3sin(2θ)=2cos(2θ) is θ=(0.58800…)/2+(pin)/2