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

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

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Popular >

cos(θ)cos(θ)=cot(θ)

Solution

cos (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

cos (θ) cos (θ) = cot (θ)

Subtract

cot (θ)

from both sides

cos^{2} (θ) - cot (θ) = 0

Express with sin, cos

cos^{2} (θ) - cot (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

- cos (θ) + cos^{2} (θ) sin (θ) = 0

Factor

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

- cos (θ) + cos^{2} (θ) sin (θ)

Apply exponent rule:

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

sin (θ) cos^{2} (θ) = cos (θ) cos (θ)

= - cos (θ) + cos (θ) cos (θ)

Factor out common term

cos (θ)

= cos (θ) (- 1 + sin (θ) cos (θ))

cos (θ) (- 1 + sin (θ) cos (θ)) = 0

Solving each part separately

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

- 1 + sin (θ) cos (θ) = 0 :

No Solution

- 1 + sin (θ) cos (θ) = 0

Rewrite using trig identities

- 1 + sin (θ) cos (θ)

Use the Double Angle identity:

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

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

= - 1 + \frac{sin ( 2 θ )}{2}

- 1 + \frac{sin ( 2 θ )}{2} = 0

Move

1

to the right side

- 1 + \frac{sin ( 2 θ )}{2} = 0

Add

1

to both sides

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

Simplify

\frac{sin ( 2 θ )}{2} = 1

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{sin ( 2 θ )}{2} = 1

Multiply both sides by

2

\frac{2 sin ( 2 θ )}{2} = 1 \cdot 2

Simplify

sin (2 θ) = 2

sin (2 θ) = 2

- 1 \leq sin (x) \leq 1

No Solution

Combine all the solutions

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

Popular Examples

3sin(θ)-cos(2θ)=1

3 sin (θ) - cos (2 θ) = 1

sin(θ)= 28/53 ,sin(2θ)

sin (θ) = \frac{28}{53}, sin (2 θ)

arctan(x)=5

arctan (x) = 5^{\circ}

tan(t)=5

tan (t) = 5

tan(t)=6

tan (t) = 6

Frequently Asked Questions (FAQ)

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