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

The general solution for 2sin(x)=(sin(x))/(cos(x)) is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin

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2sin(x)=(sin(x))/(cos(x))

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

x = 2 πn, x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Solution steps

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

Subtract

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

from both sides

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

Simplify

2 sin (x) - \frac{sin ( x )}{cos ( x )} : \frac{2 sin ( x ) cos ( x ) - sin ( x )}{cos ( x )}

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

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

2 sin (x) cos (x) - sin (x) = 0

Factor

2 sin (x) cos (x) - sin (x) : sin (x) (2 cos (x) - 1)

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

Solving each part separately

sin (x) = 0 or 2 cos (x) - 1 = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

2 cos (x) - 1 = 0 : x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

Combine all the solutions

x = 2 πn, x = π + 2 πn, x = \frac{π}{3} + 2 πn, x = \frac{5 π}{3} + 2 πn

6 sin^{2} (x) = 5 + cos (x), 18 0^{\circ} \leq x \leq 54 0^{\circ}

\frac{cos ( x )}{2} = \frac{3}{2}

2 sin (x) cos (2 x) - cos (2 x) = 0

- 2 = cos (x)

2 cot (2 x) + 2 = 0

What is the general solution for 2sin(x)=(sin(x))/(cos(x)) ?
The general solution for 2sin(x)=(sin(x))/(cos(x)) is x=2pin,x=pi+2pin,x= pi/3+2pin,x=(5pi)/3+2pin