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

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

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-3sin(2x)=5cos(2x)

- 3 sin (2 x) = 5 cos (2 x)

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

Degrees

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

Solution steps

- 3 sin (2 x) = 5 cos (2 x)

Rewrite using trig identities

- 3 sin (2 x) = 5 cos (2 x)

Divide both sides by

cos (2 x), cos (2 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

- 3 tan (2 x) = 5

- 3 tan (2 x) = 5

Divide both sides by

- 3

- 3 tan (2 x) = 5

Divide both sides by

- 3

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

2 x = arctan (- \frac{5}{3}) + πn

2 x = arctan (- \frac{5}{3}) + πn

Solve

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

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

Divide both sides by

2

2 x = - arctan (\frac{5}{3}) + πn

Divide both sides by

2

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

Simplify

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

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

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

Show solutions in decimal form

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

3 cot (x) - 1 = 0, 0 \leq x \leq 2 π

0.001 = \frac{sin ( θ ) \cdot 1 2 ^{2}}{9.81}

2 - 5 cos (x) = 0

4 cos (2 x) - 2 = 0

3 - 2 sin^{2} (\frac{x}{2}) = 2 sin^{2} (x)

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