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tan(x)-csc(2x)=0

Solution

tan (x) - csc (2 x) = 0

Solution

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Degrees

x = 22 5^{\circ} + 36 0^{\circ} n, x = 31 5^{\circ} + 36 0^{\circ} n, x = 4 5^{\circ} + 36 0^{\circ} n, x = 13 5^{\circ} + 36 0^{\circ} n

Solution steps

tan (x) - csc (2 x) = 0

Express with sin, cos

- csc (2 x) + tan (x)

Use the basic trigonometric identity:

csc (x) = \frac{1}{sin ( x )}

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

Use the basic trigonometric identity:

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

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

Simplify

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

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

Least Common Multiplier of

sin (2 x), cos (x) : sin (2 x) cos (x)

sin (2 x), cos (x)

Lowest Common Multiplier (LCM)

Compute an expression comprised of factors that appear either in

sin (2 x)

cos (x)

= sin (2 x) cos (x)

Adjust Fractions based on the LCM

Multiply each numerator by the same amount needed to multiply its

corresponding denominator to turn it into the LCM

sin (2 x) cos (x)

For

\frac{1}{sin ( 2 x )} :

multiply the denominator and numerator by

cos (x)

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

For

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

multiply the denominator and numerator by

sin (2 x)

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

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

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

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

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

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

Rewrite using trig identities

- cos (x) + sin (2 x) sin (x)

Use the Double Angle identity:

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

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

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

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

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

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

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

Factor

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

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

Factor out common term

cos (x)

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

Factor

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

2 sin^{2} (x) - 1

Rewrite

2 sin^{2} (x) - 1

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

2 sin^{2} (x) - 1

Apply radical rule:

a = (a)^{2}

2 = (2)^{2}

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

Rewrite

1

1^{2}

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

Apply exponent rule:

a^{m} b^{m} = (ab)^{m}

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

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

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

Apply Difference of Two Squares Formula:

x^{2} - y^{2} = (x + y) (x - y)

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

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

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

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

Solving each part separately

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

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

cos (x) = 0

General solutions for

cos (x) = 0

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

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

2 sin (x) + 1 = 0 : x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 sin (x) + 1 = 0

Move

1

to the right side

2 sin (x) + 1 = 0

Subtract

1

from both sides

2 sin (x) + 1 - 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

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

Simplify

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

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

Cancel the common factor:

2

= sin (x)

Simplify

\frac{- 1}{2} : - \frac{2}{2}

\frac{- 1}{2}

Apply the fraction rule:

\frac{- a}{b} = - \frac{a}{b}

= - \frac{1}{2}

Rationalize

- \frac{1}{2} : - \frac{2}{2}

- \frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= - \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= - \frac{2}{2}

= - \frac{2}{2}

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

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

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

General solutions for

sin (x) = - \frac{2}{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{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn

2 sin (x) - 1 = 0 : x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

2 sin (x) - 1 = 0

Move

1

to the right side

2 sin (x) - 1 = 0

Add

1

to both sides

2 sin (x) - 1 + 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

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

Simplify

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

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

Cancel the common factor:

2

= sin (x)

Simplify

\frac{1}{2} : \frac{2}{2}

\frac{1}{2}

Multiply by the conjugate

\frac{2}{2}

= \frac{1 \cdot 2}{2 2}

1 \cdot 2 = 2

22 = 2

22

Apply radical rule:

a a = a

22 = 2

= 2

= \frac{2}{2}

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

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

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

General solutions for

sin (x) = \frac{2}{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{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Combine all the solutions

x = \frac{π}{2} + 2 πn, x = \frac{3 π}{2} + 2 πn, x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

Since the equation is undefined for:

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

x = \frac{5 π}{4} + 2 πn, x = \frac{7 π}{4} + 2 πn, x = \frac{π}{4} + 2 πn, x = \frac{3 π}{4} + 2 πn

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