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

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

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Popular >

2sin(2θ)=2cos(θ)

Solution

2 sin (2 θ) = 2 cos (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 sin (2 θ) = 2 cos (θ)

Subtract

2 cos (θ)

from both sides

2 sin (2 θ) - 2 cos (θ) = 0

Rewrite using trig identities

- 2 cos (θ) + 2 sin (2 θ)

Use the Double Angle identity:

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

= - 2 cos (θ) + 2 \cdot 2 sin (θ) cos (θ)

Simplify

= - 2 cos (θ) + 4 sin (θ) cos (θ)

- 2 cos (θ) + 4 cos (θ) sin (θ) = 0

Factor

- 2 cos (θ) + 4 cos (θ) sin (θ) : 2 cos (θ) (2 sin (θ) - 1)

- 2 cos (θ) + 4 cos (θ) sin (θ)

Rewrite

4

2 \cdot 2

= - 2 cos (θ) + 2 \cdot 2 sin (θ) cos (θ)

Factor out common term

2 cos (θ)

= 2 cos (θ) (- 1 + 2 sin (θ))

2 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

1+sqrt(3)tan((2x-pi)/3)=0

1 + 3 tan (\frac{2 x - π}{3}) = 0

tan(3x)=tan(pix)

tan (3 x) = tan (π x)

sin^2(2x)=0.5

sin^{2} (2 x) = 0.5

sin(x)=-1/2 sqrt(2)

sin (x) = - \frac{1}{2} 2

(6cos(x)+5sin(x))^2+11sin^2(x)=66

(6 cos (x) + 5 sin (x))^{2} + 11 sin^{2} (x) = 66

Frequently Asked Questions (FAQ)

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