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

The general solution for -3sin(2θ)=5cos(2θ) is θ=-(1.03037…)/2+(pin)/2

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-3sin(2θ)=5cos(2θ)

- 3 sin (2 θ) = 5 cos (2 θ)

θ = - \frac{1.03037 \dots}{2} + \frac{πn}{2}

Solution steps

- 3 sin (2 θ) = 5 cos (2 θ)

Rewrite using trig identities

- 3 tan (2 θ) = 5

Divide both sides by

- 3

tan (2 θ) = - \frac{5}{3}

Apply trig inverse properties

2 θ = arctan (- \frac{5}{3}) + πn

Solve

2 θ = arctan (- \frac{5}{3}) + πn : θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{πn}{2}

θ = - \frac{arctan ( \frac{5}{3} )}{2} + \frac{πn}{2}

Show solutions in decimal form

θ = - \frac{1.03037 \dots}{2} + \frac{πn}{2}

(tan^{2} (θ) - 4) (2 cos (θ) + 1) = 0

cot (t) = - 1

cot^{2} (4 x) = 3

cos (x) tan (x) = cos (x)

tan (\frac{θ}{2} - \frac{π}{6}) = - 1

What is the general solution for -3sin(2θ)=5cos(2θ) ?
The general solution for -3sin(2θ)=5cos(2θ) is θ=-(1.03037…)/2+(pin)/2