What is the general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 ?

The general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 is θ=2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sin(2θ)-cos(θ)-2sin(θ)+1=0

Solution

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

Solution

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

Degrees

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

Solution steps

sin (2 θ) - cos (θ) - 2 sin (θ) + 1 = 0

Rewrite using trig identities

1 - cos (θ) + sin (2 θ) - 2 sin (θ)

Use the Double Angle identity:

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

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

1 - cos (θ) - 2 sin (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

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

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

Factor out

2 sin (θ)

from

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

2 sin (θ) cos (θ) - 2 sin (θ)

Factor out common term

2 sin (θ)

= 2 sin (θ) (cos (θ) - 1)

Factor out

- 1

from

- cos (θ) + 1 : - (cos (θ) - 1)

- cos (θ) + 1

Factor out common term

- 1

= - (cos (θ) - 1)

= 2 sin (θ) (cos (θ) - 1) - (cos (θ) - 1)

Factor out common term

cos (θ) - 1

= (cos (θ) - 1) (2 sin (θ) - 1)

(cos (θ) - 1) (2 sin (θ) - 1) = 0

Solving each part separately

cos (θ) - 1 = 0 or 2 sin (θ) - 1 = 0

cos (θ) - 1 = 0 : θ = 2 πn

cos (θ) - 1 = 0

Move

1

to the right side

cos (θ) - 1 = 0

Add

1

to both sides

cos (θ) - 1 + 1 = 0 + 1

Simplify

cos (θ) = 1

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

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

Popular Examples

2cos(x)+1=0-2,pi<= x<= 2pi

2 cos (x) + 1 = 0 - 2, π \leq x \leq 2 π

cos^3(x)=-cos(x)

cos^{3} (x) = - cos (x)

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

csc (2 x) = 2 sin (2 x), 0 \leq x \leq 2 π

cos(x)=-12/13

cos (x) = - \frac{12}{13}

solvefor x,z=arcsin(x/y)solve for

x, z = arcsin (\frac{x}{y})

Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 ?
The general solution for sin(2θ)-cos(θ)-2sin(θ)+1=0 is θ=2pin,θ= pi/6+2pin,θ=(5pi)/6+2pin