What is the general solution for (1+(2sin(x)))/((cos(x)))=0 ?

The general solution for (1+(2sin(x)))/((cos(x)))=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin

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(1+(2sin(x)))/((cos(x)))=0

\frac{1 + ( 2 sin ( x ) )}{( cos ( x ) )} = 0

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{1 + ( 2 sin ( x ) )}{( cos ( x ) )} = 0

\frac{f ( x )}{g ( x )} = 0 \Rightarrow f (x) = 0

1 + 2 sin (x) = 0

Move

1

to the right side

1 + 2 sin (x) = 0

Subtract

1

from both sides

1 + 2 sin (x) - 1 = 0 - 1

Simplify

2 sin (x) = - 1

2 sin (x) = - 1

Divide both sides by

2

2 sin (x) = - 1

Divide both sides by

2

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

Simplify

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

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

General solutions for

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

sin (x)

periodicity table with

2 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} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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

\frac{cot ^{2} ( x )}{2} = 3

sin (x - 3 0^{\circ}) = 0.7

\leq 3 \cdot cos (x) - sin (x) = 0

tan^{2} (x) + sec^{2} (x) = 3

What is the general solution for (1+(2sin(x)))/((cos(x)))=0 ?
The general solution for (1+(2sin(x)))/((cos(x)))=0 is x=(7pi)/6+2pin,x=(11pi)/6+2pin