What is the general solution for 2cos(θ)=cot(θ) ?

The general solution for 2cos(θ)=cot(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

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Popular >

2cos(θ)=cot(θ)

Solution

2 cos (θ) = cot (θ)

Solution

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Degrees

θ = 9 0^{\circ} + 36 0^{\circ} n, θ = 27 0^{\circ} + 36 0^{\circ} n, θ = 3 0^{\circ} + 36 0^{\circ} n, θ = 15 0^{\circ} + 36 0^{\circ} n

Solution steps

2 cos (θ) = cot (θ)

Subtract

cot (θ)

from both sides

2 cos (θ) - cot (θ) = 0

Express with sin, cos

- cot (θ) + 2 cos (θ)

Use the basic trigonometric identity:

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

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

Simplify

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

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

Convert element to fraction:

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

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

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

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

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

- cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

- cos (θ) + 2 cos (θ) sin (θ) : cos (θ) (2 sin (θ) - 1)

- cos (θ) + 2 cos (θ) sin (θ)

Factor out common term

cos (θ)

= cos (θ) (- 1 + 2 sin (θ))

cos (θ) (2 sin (θ) - 1) = 0

Solving each part separately

cos (θ) = 0 or 2 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

2 sin (θ) - 1 = 0 : θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

2 sin (θ) - 1 = 0

Move

1

to the right side

2 sin (θ) - 1 = 0

Add

1

to both sides

2 sin (θ) - 1 + 1 = 0 + 1

Simplify

2 sin (θ) = 1

2 sin (θ) = 1

Divide both sides by

2

2 sin (θ) = 1

Divide both sides by

2

\frac{2 sin ( θ )}{2} = \frac{1}{2}

Simplify

sin (θ) = \frac{1}{2}

sin (θ) = \frac{1}{2}

General solutions for

sin (θ) = \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}

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

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

Combine all the solutions

θ = \frac{π}{2} + 2 πn, θ = \frac{3 π}{2} + 2 πn, θ = \frac{π}{6} + 2 πn, θ = \frac{5 π}{6} + 2 πn

Popular Examples

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tan^{2} (x) = \frac{1}{3}, 0^{\circ} \leq x \leq 72 0^{\circ}

sin(2x)=1,0<= x<= 2pi

sin (2 x) = 1, 0 \leq x \leq 2 π

cos^2(z)=-1

cos^{2} (z) = - 1

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cosh (x) = \frac{6}{2}

tan(x)-sec(x)=3,0<= x<2pi

tan (x) - sec (x) = 3, 0 \leq x < 2 π

Frequently Asked Questions (FAQ)

What is the general solution for 2cos(θ)=cot(θ) ?
The general solution for 2cos(θ)=cot(θ) is θ= pi/2+2pin,θ=(3pi)/2+2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin