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

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

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cos(θ)sin(θ)=sin(2θ)

cos (θ) sin (θ) = sin (2 θ)

θ = πn, θ = \frac{π}{2} + πn

Solution steps

cos (θ) sin (θ) = sin (2 θ)

Subtract

sin (2 θ)

from both sides

cos (θ) sin (θ) - sin (2 θ) = 0

Rewrite using trig identities

- \frac{sin ( 2 θ )}{2} = 0

Multiply both sides by

2

- sin (2 θ) = 0

Divide both sides by

- 1

sin (2 θ) = 0

General solutions for

sin (2 θ) = 0

2 θ = 0 + 2 πn, 2 θ = π + 2 πn

Solve

2 θ = 0 + 2 πn : θ = πn

Solve

2 θ = π + 2 πn : θ = \frac{π}{2} + πn

θ = πn, θ = \frac{π}{2} + πn

4 sin (x) cos (x) = - 3 cos (x)

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

tan (θ) = \frac{1.0667}{0.5}, θ

cos^{2} (x) + cos (x) - 1 = 0

2 = sec (x)

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