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

The general solution for sin(2θ)=cos(θ),2 is No Solution for θ\in\mathbb{R}

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Popular >

sin(2θ)=cos(θ),2

Solution

sin (2 θ) = cos (θ), 2

Solution

No Solution for θ \in R

Solution steps

sin (2 θ) = cos (θ), 2

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, 2 :

No Solution

cos (θ) = 0, 2

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

2

No Solution

2 sin (θ) - 1 = 0, 2 :

No Solution

2 sin (θ) - 1 = 0, 2

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

2

No Solution

Combine all the solutions

No Solution for θ \in R

Popular Examples

(1+cos^2(x))/(sin^2(x))=1

\frac{1 + cos ^{2} ( x )}{sin ^{2} ( x )} = 1

(sin^2(θ)-2sin(θ)+1)/(sin(θ)-1)=-1

\frac{sin ^{2} ( θ ) - 2 sin ( θ ) + 1}{sin ( θ ) - 1} = - 1

solvefor w,100cos(10w)=50 solve for

w, 100 cos (10 w) = 50

cos(θ)= 8/25

cos (θ) = \frac{8}{25}

sin(2x)=pi

sin (2 x) = π

Frequently Asked Questions (FAQ)

What is the general solution for sin(2θ)=cos(θ),2 ?
The general solution for sin(2θ)=cos(θ),2 is No Solution for θ\in\mathbb{R}