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

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

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(1-cos(2x))/(sin(2x))=sqrt(3)

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

x = \frac{π}{3} + πn

Solution steps

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

Subtract

3

from both sides

\frac{1 - cos ( 2 x )}{sin ( 2 x )} - 3 = 0

Simplify

\frac{1 - cos ( 2 x )}{sin ( 2 x )} - 3 : \frac{1 - cos ( 2 x ) - 3 sin ( 2 x )}{sin ( 2 x )}

\frac{1 - cos ( 2 x ) - 3 sin ( 2 x )}{sin ( 2 x )} = 0

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

1 - cos (2 x) - 3 sin (2 x) = 0

Rewrite using trig identities

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

Factor

2 sin^{2} (x) - 2 cos (x) sin (x) 3 : 2 sin (x) (sin (x) - 3 cos (x))

2 sin (x) (sin (x) - 3 cos (x)) = 0

Solving each part separately

sin (x) = 0 or sin (x) - 3 cos (x) = 0

sin (x) = 0 : x = 2 πn, x = π + 2 πn

sin (x) - 3 cos (x) = 0 : x = \frac{π}{3} + πn

Combine all the solutions

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

Since the equation is undefined for:

2 πn, π + 2 πn

x = \frac{π}{3} + πn

2 sin (- 3 x - \frac{π}{4}) = 2, 0 \leq x \leq 2 π

9 sin^{2} (θ) - 9 sin (θ) - 2 = - 6 sin (θ)

5 cot (θ) + 3 = 0

cos (x) + 1 = 1

2 sin (5 x) + 3 = 0

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