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Popular >

((1-cos(x))(csc^2(x)-4))/(1+cos(x))=0

Solution

\frac{( 1 - cos ( x ) ) ( csc ^{2} ( x ) - 4 )}{1 + cos ( x )} = 0

Solution

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Degrees

x = 3 0^{\circ} + 36 0^{\circ} n, x = 15 0^{\circ} + 36 0^{\circ} n, x = 21 0^{\circ} + 36 0^{\circ} n, x = 33 0^{\circ} + 36 0^{\circ} n

Solution steps

\frac{( 1 - cos ( x ) ) ( csc ^{2} ( x ) - 4 )}{1 + cos ( x )} = 0

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

(1 - cos (x)) (csc^{2} (x) - 4) = 0

Solving each part separately

1 - cos (x) = 0 or csc^{2} (x) - 4 = 0

1 - cos (x) = 0 : x = 2 πn

1 - cos (x) = 0

Move

1

to the right side

1 - cos (x) = 0

Subtract

1

from both sides

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

Simplify

- cos (x) = - 1

- cos (x) = - 1

Divide both sides by

- 1

- cos (x) = - 1

Divide both sides by

- 1

\frac{- cos ( x )}{- 1} = \frac{- 1}{- 1}

Simplify

cos (x) = 1

cos (x) = 1

General solutions for

cos (x) = 1

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

x = 0 + 2 πn

x = 0 + 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

csc^{2} (x) - 4 = 0 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

csc^{2} (x) - 4 = 0

Solve by substitution

csc^{2} (x) - 4 = 0

Let:

csc (x) = u

u^{2} - 4 = 0

u^{2} - 4 = 0 : u = 2, u = - 2

u^{2} - 4 = 0

Move

4

to the right side

u^{2} - 4 = 0

Add

4

to both sides

u^{2} - 4 + 4 = 0 + 4

Simplify

u^{2} = 4

u^{2} = 4

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

u = 4, u = - 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

- 4 = - 2

- 4

4 = 2

4

Factor the number:

4 = 2^{2}

= 2^{2}

Apply radical rule:

2^{2} = 2

= 2

= - 2

u = 2, u = - 2

Substitute back

u = csc (x)

csc (x) = 2, csc (x) = - 2

csc (x) = 2, csc (x) = - 2

csc (x) = 2 : x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = 2

General solutions for

csc (x) = 2

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 2

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

csc (x) = - 2 : x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

csc (x) = - 2

General solutions for

csc (x) = - 2

csc (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} csc (x) Undefiend 22 \frac{2 3}{3} 1 \frac{2 3}{3} 22 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} csc (x) Undefiend - 2 - 2 - \frac{2 3}{3} - 1 - \frac{2 3}{3} - 2 - 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

Combine all the solutions

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Combine all the solutions

x = 2 πn, x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

Since the equation is undefined for:

2 πn

x = \frac{π}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn, x = \frac{7 π}{6} + 2 πn, x = \frac{11 π}{6} + 2 πn

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