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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

-3sin(2θ)=5cos(2θ)

- 3 sin (2 θ) = 5 cos (2 θ)

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

Degrees

θ = - 29.51812 \dots^{\circ} + 9 0^{\circ} n

Solution steps

- 3 sin (2 θ) = 5 cos (2 θ)

Rewrite using trig identities

- 3 sin (2 θ) = 5 cos (2 θ)

Divide both sides by

cos (2 θ), cos (2 θ) \neq = 0

\frac{- 3 sin ( 2 θ )}{cos ( 2 θ )} = \frac{5 cos ( 2 θ )}{cos ( 2 θ )}

Simplify

- \frac{3 sin ( 2 θ )}{cos ( 2 θ )} = 5

Use the basic trigonometric identity:

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

- 3 tan (2 θ) = 5

- 3 tan (2 θ) = 5

Divide both sides by

- 3

- 3 tan (2 θ) = 5

Divide both sides by

- 3

\frac{- 3 tan ( 2 θ )}{- 3} = \frac{5}{- 3}

Simplify

tan (2 θ) = - \frac{5}{3}

tan (2 θ) = - \frac{5}{3}

Apply trig inverse properties

tan (2 θ) = - \frac{5}{3}

General solutions for

tan (2 θ) = - \frac{5}{3}

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

2 θ = arctan (- \frac{5}{3}) + πn

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}

2 θ = arctan (- \frac{5}{3}) + πn

Simplify

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

arctan (- \frac{5}{3}) + πn

Use the following property:

arctan (- x) = - arctan (x)

arctan (- \frac{5}{3}) = - arctan (\frac{5}{3})

= - arctan (\frac{5}{3}) + πn

2 θ = - arctan (\frac{5}{3}) + πn

Divide both sides by

2

2 θ = - arctan (\frac{5}{3}) + πn

Divide both sides by

2

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

Simplify

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

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