What is the general solution for sqrt(3)cos(1x)=-sin(1x) ?

The general solution for sqrt(3)cos(1x)=-sin(1x) is x=(2pi)/3+pin

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sqrt(3)cos(1x)=-sin(1x)

3 cos (1 x) = - sin (1 x)

x = \frac{2 π}{3} + πn

Degrees

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

Solution steps

3 cos (1 \cdot x) = - sin (1 \cdot x)

Rewrite using trig identities

3 cos (1 \cdot x) = - sin (1 \cdot x)

Divide both sides by

cos (1 x), cos (1 x) \neq = 0

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

Simplify

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

Use the basic trigonometric identity:

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

3 = - tan (x)

Switch sides

- 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

0 = 3 cos (θ)

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

sin^{2} (θ) - sin (θ) + 1 = cos^{2} (θ)

1 = l cos (x)

tan (2 x) = cos (2 x), 0 \leq x \leq π

What is the general solution for sqrt(3)cos(1x)=-sin(1x) ?
The general solution for sqrt(3)cos(1x)=-sin(1x) is x=(2pi)/3+pin