What is the general solution for sin(2θ)=cos(θ),0<= θ<2pi ?

Q: What is the general solution for sin(2θ)=cos(θ),0<= θ<2pi ?

The general solution for sin(2θ)=cos(θ),0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2 ,θ= pi/6 ,θ=(5pi)/6

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sin(2θ)=cos(θ),0<= θ<2pi

Solution

sin (2 θ) = cos (θ), 0 \leq θ < 2 π

Solution

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

Degrees

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

Solution steps

sin (2 θ) = cos (θ), 0 \leq θ < 2 π

Subtract

cos (θ)

from both sides

sin (2 θ) - cos (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, 0 \leq θ < 2 π : θ = \frac{π}{2}, θ = \frac{3 π}{2}

cos (θ) = 0, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{2}, θ = \frac{3 π}{2}

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

2 sin (θ) - 1 = 0, 0 \leq θ < 2 π

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

Solutions for the range

0 \leq θ < 2 π

θ = \frac{π}{6}, θ = \frac{5 π}{6}

Combine all the solutions

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

Popular Examples

3/2 sec(θ)=3

\frac{3}{2} sec (θ) = 3

4csc(θ)-3=0

4 csc (θ) - 3 = 0

2cos^2(θ)-10=-cos(θ)-9

2 cos^{2} (θ) - 10 = - cos (θ) - 9

1=tan(pi+c)

1 = tan (π + c)

6193cos(x)+2880cos(2x)=0

6193 cos (x) + 2880 cos (2 x) = 0

Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)=cos(θ),0<= θ<2pi ?
The general solution for sin(2θ)=cos(θ),0<= θ<2pi is θ= pi/2 ,θ=(3pi)/2 ,θ= pi/6 ,θ=(5pi)/6