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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Popular Trigonometry >

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

Solution

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

Solution

x = \frac{π}{3} + πn

Degrees

x = 6 0^{\circ} + 18 0^{\circ} n

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 )}{sin ( 2 x )} - 3

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

\frac{a}{c} \pm \frac{b}{c} = \frac{a \pm b}{c}

= \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

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

Use the Double Angle identity:

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

= 1 - cos (2 x) - 3 \cdot 2 sin (x) cos (x)

Use the Double Angle identity:

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

= 1 - (1 - 2 sin^{2} (x)) - 23 cos (x) sin (x)

Simplify

1 - (1 - 2 sin^{2} (x)) - 23 cos (x) sin (x) : 2 sin^{2} (x) - 23 cos (x) sin (x)

1 - (1 - 2 sin^{2} (x)) - 23 cos (x) sin (x)

- (1 - 2 sin^{2} (x)) : - 1 + 2 sin^{2} (x)

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

Distribute parentheses

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

Apply minus-plus rules

- (- a) = a, - (a) = - a

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

= 1 - 1 + 2 sin^{2} (x) - 23 cos (x) sin (x)

1 - 1 = 0

= 2 sin^{2} (x) - 23 cos (x) sin (x)

= 2 sin^{2} (x) - 23 cos (x) sin (x)

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

Apply exponent rule:

a^{b + c} = a^{b} a^{c}

sin^{2} (x) = sin (x) sin (x)

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

Rewrite as

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

Factor out common term

2 sin (x)

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

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) = 0

General solutions for

sin (x) = 0

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 = 0 + 2 πn, x = π + 2 πn

x = 0 + 2 πn, x = π + 2 πn

Solve

x = 0 + 2 πn : x = 2 πn

x = 0 + 2 πn

0 + 2 πn = 2 πn

x = 2 πn

x = 2 πn, x = π + 2 πn

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

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

Rewrite using trig identities

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

Divide both sides by

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

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

Simplify

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

Use the basic trigonometric identity:

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

tan (x) - 3 = 0

tan (x) - 3 = 0

Move

3

to the right side

tan (x) - 3 = 0

Add

3

to both sides

tan (x) - 3 + 3 = 0 + 3

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{π}{3} + πn

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

Graph

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Frequently Asked Questions (FAQ)

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