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

The general solution for 2cos(θ)+sin(2θ)=0 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 >

2cos(θ)+sin(2θ)=0

2 cos (θ) + sin (2 θ) = 0

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

2 cos (θ) + sin (2 θ) = 0

Rewrite using trig identities

sin (2 θ) + 2 cos (θ)

Use the Double Angle identity:

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

= 2 sin (θ) cos (θ) + 2 cos (θ)

2 cos (θ) + 2 cos (θ) sin (θ) = 0

Factor

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

2 cos (θ) + 2 cos (θ) sin (θ)

Factor out common term

2 cos (θ)

= 2 cos (θ) (1 + sin (θ))

2 cos (θ) (sin (θ) + 1) = 0

Solving each part separately

cos (θ) = 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

Combine all the solutions

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

tan (x) = \frac{6 \cdot 3}{6}

cos^{2} (x) + sin^{2} (x) - 2 sin (x) = 0

sin (x) = cos^{2} (x) - sin^{2} (x) - 1

2 + cos (2 x) = - 3 cos (x)

tan (\frac{3 x}{2} + \frac{π}{2}) = 1

What is the general solution for 2cos(θ)+sin(2θ)=0 ?
The general solution for 2cos(θ)+sin(2θ)=0 is θ= pi/2+2pin,θ=(3pi)/2+2pin