What is the general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) ?

The general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) is No Solution for θ\in\mathbb{R}

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cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ))

Solution

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

Solution

No Solution for θ \in R

Solution steps

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

Subtract

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

from both sides

cot (θ) - \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} = 0

Simplify

cot (θ) - \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} : \frac{2 cot ( θ ) sin ( θ ) cos ( θ ) - 1 - cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

cot (θ) - \frac{1 + cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

Convert element to fraction:

cot (θ) = \frac{cot ( θ ) 2 sin ( θ ) cos ( θ )}{2 sin ( θ ) cos ( θ )}

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

Since the denominators are equal, combine the fractions:

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

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

Expand

cot (θ) \cdot 2 sin (θ) cos (θ) - (1 + cos^{2} (θ)) : cot (θ) \cdot 2 sin (θ) cos (θ) - 1 - cos^{2} (θ)

cot (θ) \cdot 2 sin (θ) cos (θ) - (1 + cos^{2} (θ))

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

- (1 + cos^{2} (θ)) : - 1 - cos^{2} (θ)

- (1 + cos^{2} (θ))

Distribute parentheses

= - (1) - (cos^{2} (θ))

Apply minus-plus rules

+ (- a) = - a

= - 1 - cos^{2} (θ)

= cot (θ) \cdot 2 sin (θ) cos (θ) - 1 - cos^{2} (θ)

= \frac{2 cot ( θ ) sin ( θ ) cos ( θ ) - 1 - cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )}

\frac{2 cot ( θ ) sin ( θ ) cos ( θ ) - 1 - cos ^{2} ( θ )}{2 sin ( θ ) cos ( θ )} = 0

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

2 cot (θ) sin (θ) cos (θ) - 1 - cos^{2} (θ) = 0

Rewrite using trig identities

- 1 - cos^{2} (θ) + 2 cos (θ) cot (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

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

2 cos (θ) \frac{cos ( θ )}{sin ( θ )} sin (θ)

Multiply fractions:

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

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

Cancel the common factor:

sin (θ)

= cos (θ) \cdot 2 cos (θ)

Apply exponent rule:

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

cos (θ) cos (θ) = cos^{1 + 1} (θ)

= 2 cos^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= 2 cos^{2} (θ)

= - 1 - cos^{2} (θ) + 2 cos^{2} (θ)

Add similar elements:

- cos^{2} (θ) + 2 cos^{2} (θ) = cos^{2} (θ)

= - 1 + cos^{2} (θ)

= - 1 + cos^{2} (θ)

- 1 + cos^{2} (θ) = 0

Solve by substitution

- 1 + cos^{2} (θ) = 0

Let:

cos (θ) = u

- 1 + u^{2} = 0

- 1 + u^{2} = 0 : u = 1, u = - 1

- 1 + u^{2} = 0

Move

1

to the right side

- 1 + u^{2} = 0

Add

1

to both sides

- 1 + u^{2} + 1 = 0 + 1

Simplify

u^{2} = 1

u^{2} = 1

For

x^{2} = f (a)

the solutions are

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

u = 1, u = - 1

1 = 1

1

Apply rule

1 = 1

= 1

- 1 = - 1

- 1

Apply rule

1 = 1

= - 1

u = 1, u = - 1

Substitute back

u = cos (θ)

cos (θ) = 1, cos (θ) = - 1

cos (θ) = 1, cos (θ) = - 1

cos (θ) = 1 : θ = 2 πn

cos (θ) = 1

General solutions for

cos (θ) = 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}

θ = 0 + 2 πn

θ = 0 + 2 πn

Solve

θ = 0 + 2 πn : θ = 2 πn

θ = 0 + 2 πn

0 + 2 πn = 2 πn

θ = 2 πn

θ = 2 πn

cos (θ) = - 1 : θ = π + 2 πn

cos (θ) = - 1

General solutions for

cos (θ) = - 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}

θ = π + 2 πn

θ = π + 2 πn

Combine all the solutions

θ = 2 πn, θ = π + 2 πn

Since the equation is undefined for:

2 πn, π + 2 πn

No Solution for θ \in R

Popular Examples

cot(x)=sqrt(3)+1

cot (x) = 3 + 1

sec(3x)-csc(30)=0,(x+35)/5

sec (3 x) - csc (3 0^{\circ}) = 0, \frac{x + 35}{5}

4cos(2θ)-10cos(θ)+14=7,0<= θ<360

4 cos (2 θ) - 10 cos (θ) + 14 = 7, 0 \leq θ < 36 0^{\circ}

solvefor x,Z=sin(xy^2)solve for

x, Z = sin (x y^{2})

2tan^2(θ)=2

2 tan^{2} (θ) = 2

Frequently Asked Questions (FAQ)

What is the general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) ?
The general solution for cot(θ)=(1+cos^2(θ))/(2sin(θ)cos(θ)) is No Solution for θ\in\mathbb{R}