What is the general solution for sin(θ)cos(θ)=cot(θ) ?

The general solution for sin(θ)cos(θ)=cot(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(θ)cos(θ)=cot(θ)

Solution

sin (θ) cos (θ) = cot (θ)

Solution

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

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n

Solution steps

sin (θ) cos (θ) = cot (θ)

Subtract

cot (θ)

from both sides

sin (θ) cos (θ) - cot (θ) = 0

Express with sin, cos

- cot (θ) + cos (θ) sin (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

- \frac{cos ( θ )}{sin ( θ )} + cos (θ) sin (θ)

Convert element to fraction:

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

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

Since the denominators are equal, combine the fractions:

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

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

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

- cos (θ) + cos (θ) sin (θ) sin (θ)

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

cos (θ) sin (θ) sin (θ)

Apply exponent rule:

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

sin (θ) sin (θ) = sin^{1 + 1} (θ)

= cos (θ) sin^{1 + 1} (θ)

Add the numbers:

1 + 1 = 2

= cos (θ) sin^{2} (θ)

= - cos (θ) + sin^{2} (θ) cos (θ)

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

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

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

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

- cos (θ) + cos (θ) sin^{2} (θ) = 0

Factor

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

- cos (θ) + cos (θ) sin^{2} (θ)

Factor out common term

cos (θ)

= cos (θ) (- 1 + sin^{2} (θ))

Factor

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

sin^{2} (θ) - 1

Rewrite

1

1^{2}

= sin^{2} (θ) - 1^{2}

Apply Difference of Two Squares Formula:

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

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

= (sin (θ) + 1) (sin (θ) - 1)

= cos (θ) (sin (θ) + 1) (sin (θ) - 1)

cos (θ) (sin (θ) + 1) (sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or sin (θ) + 1 = 0 or sin (θ) - 1 = 0

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

cos (θ) = 0

General solutions for

cos (θ) = 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}

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

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

sin (θ) + 1 = 0 : θ = \frac{3 π}{2} + 2 πn

sin (θ) + 1 = 0

Move

1

to the right side

sin (θ) + 1 = 0

Subtract

1

from both sides

sin (θ) + 1 - 1 = 0 - 1

Simplify

sin (θ) = - 1

sin (θ) = - 1

General solutions for

sin (θ) = - 1

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}

θ = \frac{3 π}{2} + 2 πn

θ = \frac{3 π}{2} + 2 πn

sin (θ) - 1 = 0 : θ = \frac{π}{2} + 2 πn

sin (θ) - 1 = 0

Move

1

to the right side

sin (θ) - 1 = 0

Add

1

to both sides

sin (θ) - 1 + 1 = 0 + 1

Simplify

sin (θ) = 1

sin (θ) = 1

General solutions for

sin (θ) = 1

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}

θ = \frac{π}{2} + 2 πn

θ = \frac{π}{2} + 2 πn

Combine all the solutions

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

Popular Examples

cot(θ)=-4/3

cot (θ) = - \frac{4}{3}

tan(x)= 6/10

tan (x) = \frac{6}{10}

tan(A)=34

tan (A) = 34

sqrt(2)sec(x)=2

2 sec (x) = 2

-2cos(2θ)+9cos(θ)+4=cos(θ)+1

- 2 cos (2 θ) + 9 cos (θ) + 4 = cos (θ) + 1

Frequently Asked Questions (FAQ)

What is the general solution for sin(θ)cos(θ)=cot(θ) ?
The general solution for sin(θ)cos(θ)=cot(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin