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}

Solution steps

3 sin (2 θ) = 2 cos (2 θ)

Rewrite using trig identities

3 tan (2 θ) = 2

Divide both sides by

3

tan (2 θ) = \frac{2}{3}

Apply trig inverse properties

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}

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