What is the general solution for 4sin(θ)=-2cos(θ) ?

The general solution for 4sin(θ)=-2cos(θ) is θ=-0.46364…+pin

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4sin(θ)=-2cos(θ)

4 sin (θ) = - 2 cos (θ)

θ = - 0.46364 \dots + πn

Degrees

θ = - 26.56505 \dots^{\circ} + 18 0^{\circ} n

Solution steps

4 sin (θ) = - 2 cos (θ)

Rewrite using trig identities

4 sin (θ) = - 2 cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

\frac{4 sin ( θ )}{cos ( θ )} = \frac{- 2 cos ( θ )}{cos ( θ )}

Simplify

\frac{4 sin ( θ )}{cos ( θ )} = - 2

Use the basic trigonometric identity:

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

4 tan (θ) = - 2

4 tan (θ) = - 2

Divide both sides by

4

4 tan (θ) = - 2

Divide both sides by

4

\frac{4 tan ( θ )}{4} = \frac{- 2}{4}

Simplify

\frac{4 tan ( θ )}{4} = \frac{- 2}{4}

Simplify

\frac{4 tan ( θ )}{4} : tan (θ)

\frac{4 tan ( θ )}{4}

Divide the numbers:

\frac{4}{4} = 1

= tan (θ)

Simplify

\frac{- 2}{4} : - \frac{1}{2}

\frac{- 2}{4}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{2}{4}

Cancel the common factor:

2

= - \frac{1}{2}

tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{1}{2}

tan (θ) = - \frac{1}{2}

Apply trig inverse properties

tan (θ) = - \frac{1}{2}

General solutions for

tan (θ) = - \frac{1}{2}

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

θ = arctan (- \frac{1}{2}) + πn

θ = arctan (- \frac{1}{2}) + πn

Show solutions in decimal form

θ = - 0.46364 \dots + πn

cot (2 x) = - 1, (0, 2 π)

(sin (x) + 2) (tan (x) - 2) = 0

sec^{2} (x) + 4 sec (x) - 12 = 0

sin (θ) = - 1, 0^{\circ} \leq θ < 36 0^{\circ}

3 cos^{2} (x) + sin (x) + 1 = 0

What is the general solution for 4sin(θ)=-2cos(θ) ?
The general solution for 4sin(θ)=-2cos(θ) is θ=-0.46364…+pin