What is the general solution for cos(x)=(2tan(x))/(1+tan^2(x)) ?

The general solution for cos(x)=(2tan(x))/(1+tan^2(x)) is x= pi/6+2pin,x=(5pi)/6+2pin

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cos(x)=(2tan(x))/(1+tan^2(x))

Solution

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

Solution

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

Degrees

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

Solution steps

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

Subtract

\frac{2 tan ( x )}{1 + tan ^{2} ( x )}

from both sides

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

cos (x) (1 + tan^{2} (x)) - 2 tan (x) = 0

Express with sin, cos

(1 + tan^{2} (x)) cos (x) - 2 tan (x)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

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

Apply exponent rule:

(\frac{a}{b})^{c} = \frac{a ^{c}}{b ^{c}}

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

Join

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

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

Convert element to fraction:

1 = \frac{1 cos ^{2} ( x )}{cos ^{2} ( x )}

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

Since the denominators are equal, combine the fractions:

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

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

Multiply:

1 \cdot cos^{2} (x) = cos^{2} (x)

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

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

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

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

Multiply fractions:

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

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

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

Multiply

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

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

Multiply fractions:

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

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

Cancel the common factor:

cos (x)

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

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

Since the denominators are equal, combine the fractions:

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

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

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

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

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

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

Rewrite using trig identities

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

Use the Pythagorean identity:

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

= - 2 sin (x) + 1

- 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

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

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

General solutions for

sin (x) = \frac{1}{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}{6} + 2 πn, x = \frac{5 π}{6} + 2 πn

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

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cos(2x)-cos(6x)-sin(4x)=0

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cot (θ) = \frac{15}{8}

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solvefor u,sin(u)=x solve for

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Frequently Asked Questions (FAQ)

What is the general solution for cos(x)=(2tan(x))/(1+tan^2(x)) ?
The general solution for cos(x)=(2tan(x))/(1+tan^2(x)) is x= pi/6+2pin,x=(5pi)/6+2pin