What is the general solution for 2sqrt(3)cos(z)+2sin(z)=0 ?

The general solution for 2sqrt(3)cos(z)+2sin(z)=0 is z=(2pi)/3+pin

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2sqrt(3)cos(z)+2sin(z)=0

23 cos (z) + 2 sin (z) = 0

z = \frac{2 π}{3} + πn

Degrees

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

Solution steps

23 cos (z) + 2 sin (z) = 0

Rewrite using trig identities

23 cos (z) + 2 sin (z) = 0

Divide both sides by

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

\frac{2 3 cos ( z ) + 2 sin ( z )}{cos ( z )} = \frac{0}{cos ( z )}

Simplify

23 + \frac{2 sin ( z )}{cos ( z )} = 0

Use the basic trigonometric identity:

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

23 + 2 tan (z) = 0

23 + 2 tan (z) = 0

Move

23

to the right side

23 + 2 tan (z) = 0

Subtract

23

from both sides

23 + 2 tan (z) - 23 = 0 - 23

Simplify

2 tan (z) = - 23

2 tan (z) = - 23

Divide both sides by

2

2 tan (z) = - 23

Divide both sides by

2

\frac{2 tan ( z )}{2} = \frac{- 2 3}{2}

Simplify

\frac{2 tan ( z )}{2} = \frac{- 2 3}{2}

Simplify

\frac{2 tan ( z )}{2} : tan (z)

\frac{2 tan ( z )}{2}

Divide the numbers:

\frac{2}{2} = 1

= tan (z)

Simplify

\frac{- 2 3}{2} : - 3

\frac{- 2 3}{2}

Apply the fraction rule:

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

= - \frac{2 3}{2}

Divide the numbers:

\frac{2}{2} = 1

= - 3

tan (z) = - 3

tan (z) = - 3

tan (z) = - 3

General solutions for

tan (z) = - 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}

z = \frac{2 π}{3} + πn

z = \frac{2 π}{3} + πn

\frac{cos ( x ) + cos ( 3 x )}{cos ^{2} ( x ) - sin ^{2} ( x )} = 0

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

tan (t) = \frac{- 5}{12}

\frac{( 1 - ( cos ( x ) ) ^{2} )}{( 1 + sin ( x ) )} = sin (x)

x, arctan (x) = 0

What is the general solution for 2sqrt(3)cos(z)+2sin(z)=0 ?
The general solution for 2sqrt(3)cos(z)+2sin(z)=0 is z=(2pi)/3+pin