What is the general solution for (sin(2x))/(cos(x))=4+sin(2x)*tan(x) ?

The general solution for (sin(2x))/(cos(x))=4+sin(2x)*tan(x) is No Solution for x\in\mathbb{R}

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(sin(2x))/(cos(x))=4+sin(2x)*tan(x)

Solution

\frac{sin ( 2 x )}{cos ( x )} = 4 + sin (2 x) \cdot tan (x)

Solution

No Solution for x \in R

Solution steps

\frac{sin ( 2 x )}{cos ( x )} = 4 + sin (2 x) tan (x)

Subtract

4 + sin (2 x) tan (x)

from both sides

\frac{sin ( 2 x )}{cos ( x )} - 4 - sin (2 x) tan (x) = 0

Simplify

\frac{sin ( 2 x )}{cos ( x )} - 4 - sin (2 x) tan (x) : \frac{sin ( 2 x ) - 4 cos ( x ) - sin ( 2 x ) tan ( x ) cos ( x )}{cos ( x )}

\frac{sin ( 2 x )}{cos ( x )} - 4 - sin (2 x) tan (x)

Convert element to fraction:

4 = \frac{4 cos ( x )}{cos ( x )}, sin (2 x) tan (x) = \frac{sin ( 2 x ) tan ( x ) cos ( x )}{cos ( x )}

= \frac{sin ( 2 x )}{cos ( x )} - \frac{4 cos ( x )}{cos ( x )} - \frac{sin ( 2 x ) tan ( x ) cos ( x )}{cos ( x )}

Since the denominators are equal, combine the fractions:

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

= \frac{sin ( 2 x ) - 4 cos ( x ) - sin ( 2 x ) tan ( x ) cos ( x )}{cos ( x )}

\frac{sin ( 2 x ) - 4 cos ( x ) - sin ( 2 x ) tan ( x ) cos ( x )}{cos ( x )} = 0

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

sin (2 x) - 4 cos (x) - sin (2 x) tan (x) cos (x) = 0

Rewrite using trig identities

sin (2 x) - 4 cos (x) - cos (x) sin (2 x) tan (x)

Use the Double Angle identity:

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

= 2 sin (x) cos (x) - 4 cos (x) - cos (x) \cdot 2 sin (x) cos (x) tan (x)

cos (x) \cdot 2 sin (x) cos (x) tan (x) = 2 cos^{2} (x) sin (x) tan (x)

cos (x) \cdot 2 sin (x) cos (x) tan (x)

Apply exponent rule:

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

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

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

Add the numbers:

1 + 1 = 2

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

= 2 sin (x) cos (x) - 4 cos (x) - 2 cos^{2} (x) sin (x) tan (x)

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

Factor

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

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

Apply exponent rule:

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

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

= - 4 cos (x) + 2 cos (x) sin (x) - 2 cos (x) cos (x) sin (x) tan (x)

Rewrite as

= - 2 \cdot 2 cos (x) + 2 cos (x) sin (x) - 2 cos (x) cos (x) sin (x) tan (x)

Factor out common term

2 cos (x)

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

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

Solving each part separately

cos (x) = 0 or - 2 + sin (x) - cos (x) sin (x) tan (x) = 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) - cos (x) sin (x) tan (x) = 0 :

No Solution

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

Rewrite using trig identities

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

Use the basic trigonometric identity:

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

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

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

= sin (x) sin (x)

Apply exponent rule:

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

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

= sin^{1 + 1} (x)

Add the numbers:

1 + 1 = 2

= sin^{2} (x)

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

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

Solve by substitution

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

Let:

sin (x) = u

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

- 2 + u - u^{2} = 0 : u = \frac{1}{2} - i \frac{7}{2}, u = \frac{1}{2} + i \frac{7}{2}

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

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

Quadratic Equation Formula:

For

a = - 1, b = 1, c = - 2

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

u_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) ( - 2 )}{2 ( - 1 )}

Simplify

1^{2} - 4 (- 1) (- 2) : 7 i

1^{2} - 4 (- 1) (- 2)

Apply rule

1^{a} = 1

1^{2} = 1

= 1 - 4 (- 1) (- 2)

Apply rule

- (- a) = a

= 1 - 4 \cdot 1 \cdot 2

Multiply the numbers:

4 \cdot 1 \cdot 2 = 8

= 1 - 8

Subtract the numbers:

1 - 8 = - 7

= - 7

Apply radical rule:

- a = - 1 a

- 7 = - 1 7

= - 1 7

Apply imaginary number rule:

- 1 = i

= 7 i

u_{1, 2} = \frac{- 1 \pm 7 i}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 1 + 7 i}{2 ( - 1 )}, u_{2} = \frac{- 1 - 7 i}{2 ( - 1 )}

u = \frac{- 1 + 7 i}{2 ( - 1 )} : \frac{1}{2} - i \frac{7}{2}

\frac{- 1 + 7 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 + 7 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 + 7 i}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 + 7 i}{2}

Rewrite

- \frac{- 1 + 7 i}{2}

in standard complex form:

\frac{1}{2} - \frac{7}{2} i

- \frac{- 1 + 7 i}{2}

Apply the fraction rule:

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

\frac{- 1 + 7 i}{2} = - (- \frac{1}{2}) - (\frac{7 i}{2})

= - (- \frac{1}{2}) - (\frac{7 i}{2})

Remove parentheses:

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

= \frac{1}{2} - \frac{7 i}{2}

= \frac{1}{2} - \frac{7}{2} i

u = \frac{- 1 - 7 i}{2 ( - 1 )} : \frac{1}{2} + i \frac{7}{2}

\frac{- 1 - 7 i}{2 ( - 1 )}

Remove parentheses:

(- a) = - a

= \frac{- 1 - 7 i}{- 2 \cdot 1}

Multiply the numbers:

2 \cdot 1 = 2

= \frac{- 1 - 7 i}{- 2}

Apply the fraction rule:

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

= - \frac{- 1 - 7 i}{2}

Rewrite

- \frac{- 1 - 7 i}{2}

in standard complex form:

\frac{1}{2} + \frac{7}{2} i

- \frac{- 1 - 7 i}{2}

Apply the fraction rule:

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

\frac{- 1 - 7 i}{2} = - (- \frac{1}{2}) - (- \frac{7 i}{2})

= - (- \frac{1}{2}) - (- \frac{7 i}{2})

Apply rule

- (- a) = a

= \frac{1}{2} + \frac{7 i}{2}

= \frac{1}{2} + \frac{7}{2} i

The solutions to the quadratic equation are:

u = \frac{1}{2} - i \frac{7}{2}, u = \frac{1}{2} + i \frac{7}{2}

Substitute back

u = sin (x)

sin (x) = \frac{1}{2} - i \frac{7}{2}, sin (x) = \frac{1}{2} + i \frac{7}{2}

sin (x) = \frac{1}{2} - i \frac{7}{2}, sin (x) = \frac{1}{2} + i \frac{7}{2}

sin (x) = \frac{1}{2} - i \frac{7}{2} :

No Solution

sin (x) = \frac{1}{2} - i \frac{7}{2}

No Solution

sin (x) = \frac{1}{2} + i \frac{7}{2} :

No Solution

sin (x) = \frac{1}{2} + i \frac{7}{2}

No Solution

Combine all the solutions

No Solution

Combine all the solutions

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

Since the equation is undefined for:

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

No Solution for x \in R

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Frequently Asked Questions (FAQ)

What is the general solution for (sin(2x))/(cos(x))=4+sin(2x)*tan(x) ?
The general solution for (sin(2x))/(cos(x))=4+sin(2x)*tan(x) is No Solution for x\in\mathbb{R}