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

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

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Popular >

cos(θ)=sin(2θ)

Solution

cos (θ) = sin (2 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

cos (θ) = sin (2 θ)

Subtract

sin (2 θ)

from both sides

cos (θ) - sin (2 θ) = 0

Rewrite using trig identities

cos (θ) - sin (2 θ)

Use the Double Angle identity:

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

= cos (θ) - 2 sin (θ) cos (θ)

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

solvefor θ,tan(θ)=1 solve for

θ, tan (θ) = 1

sec(x)csc(x)-2csc(x)=0

sec (x) csc (x) - 2 csc (x) = 0

csc(3θ)=6sin(3θ),0<θ<2pi

csc (3 θ) = 6 sin (3 θ), 0 < θ < 2 π

sqrt(3)tan(3θ)-1=0,0<= θ<= 2pi

3 tan (3 θ) - 1 = 0, 0 \leq θ \leq 2 π

(cot(θ)+1)(csc(θ)-1)=0

(cot (θ) + 1) (csc (θ) - 1) = 0

Frequently Asked Questions (FAQ)

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