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

The general solution for sqrt(3)sin(θ)=-cos(θ) is θ=(5pi)/6+pin

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sqrt(3)sin(θ)=-cos(θ)

3 sin (θ) = - cos (θ)

θ = \frac{5 π}{6} + πn

Degrees

θ = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

3 sin (θ) = - cos (θ)

Rewrite using trig identities

3 sin (θ) = - cos (θ)

Divide both sides by

cos (θ), cos (θ) \neq = 0

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

Simplify

\frac{3 sin ( θ )}{cos ( θ )} = - 1

Use the basic trigonometric identity:

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

3 tan (θ) = - 1

3 tan (θ) = - 1

Divide both sides by

3

3 tan (θ) = - 1

Divide both sides by

3

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} = \frac{- 1}{3}

Simplify

\frac{3 tan ( θ )}{3} : tan (θ)

\frac{3 tan ( θ )}{3}

Cancel the common factor:

3

= tan (θ)

Simplify

\frac{- 1}{3} : - \frac{3}{3}

\frac{- 1}{3}

Apply the fraction rule:

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

= - \frac{1}{3}

Rationalize

- \frac{1}{3} : - \frac{3}{3}

- \frac{1}{3}

Multiply by the conjugate

\frac{3}{3}

= - \frac{1 \cdot 3}{3 3}

1 \cdot 3 = 3

33 = 3

33

Apply radical rule:

a a = a

33 = 3

= 3

= - \frac{3}{3}

= - \frac{3}{3}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

tan (θ) = - \frac{3}{3}

General solutions for

tan (θ) = - \frac{3}{3}

tan (x)

periodicity table with

πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} tan (x) 0 \frac{3}{3} 13 \pm \infty - 3 - 1 - \frac{3}{3}

θ = \frac{5 π}{6} + πn

θ = \frac{5 π}{6} + πn

- 2 sin (x) + 1 = 0

sin (2 θ) = - \frac{3}{2}, 0 \leq θ \leq 2 π

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

3 tanh^{2} (x) = 5 sech (x) + 1

tan (θ) = \frac{5}{7}

What is the general solution for sqrt(3)sin(θ)=-cos(θ) ?
The general solution for sqrt(3)sin(θ)=-cos(θ) is θ=(5pi)/6+pin