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

The general solution for -sin(x)=sqrt(3)cos(x) is x=(2pi)/3+pin

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-sin(x)=sqrt(3)cos(x)

- sin (x) = 3 cos (x)

x = \frac{2 π}{3} + πn

Degrees

x = 12 0^{\circ} + 18 0^{\circ} n

Solution steps

- sin (x) = 3 cos (x)

Rewrite using trig identities

- sin (x) = 3 cos (x)

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

- tan (x) = 3

- tan (x) = 3

Divide both sides by

- 1

- tan (x) = 3

Divide both sides by

- 1

\frac{- tan ( x )}{- 1} = \frac{3}{- 1}

Simplify

tan (x) = - 3

tan (x) = - 3

General solutions for

tan (x) = - 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}

x = \frac{2 π}{3} + πn

x = \frac{2 π}{3} + πn

cos (2 θ) = sin (θ), 0 \leq θ < 2 π

sin (x) (1 + cos (x)) = sin (x) + cos (x)

9 0^{2} = 6 0^{2} + 7 0^{2} - 2 (60) (70) (cos (x))

sin^{2} (x) + 3 sin (x) + 7 = 5

cos (2 x) = - 0.6, - 18 0^{\circ} \leq x \leq 18 0^{\circ}

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