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

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

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

- 3 sin (x) = cos (x)

x = \frac{5 π}{6} + πn

Degrees

x = 15 0^{\circ} + 18 0^{\circ} n

Solution steps

- 3 sin (x) = cos (x)

Rewrite using trig identities

- 3 sin (x) = cos (x)

Divide both sides by

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

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

Simplify

- \frac{3 sin ( x )}{cos ( x )} = 1

Use the basic trigonometric identity:

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

- 3 tan (x) = 1

- 3 tan (x) = 1

Divide both sides by

- 3

- 3 tan (x) = 1

Divide both sides by

- 3

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

Simplify

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

Simplify

\frac{- 3 tan ( x )}{- 3} : tan (x)

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

Apply the fraction rule:

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

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

Cancel the common factor:

3

= tan (x)

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

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

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

General solutions for

tan (x) = - \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}

x = \frac{5 π}{6} + πn

x = \frac{5 π}{6} + πn

2 csc^{2} (x) = 5 - 5 cot (x)

sin (x + 0.5) = 0.92, 0 \leq x \leq 2 π

cos (4 x) = 0.8 - 0.6

5 cos^{2} (x) = 4 - 3 sin^{2} (x)

\frac{18 cos ( 2 θ ) + 2 + cos ( θ )}{2} = 0

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