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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cos(θ)=sin(2θ)

cos (θ) = sin (2 θ)

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

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) - 2 cos (θ) sin (θ) = 0

Factor

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

- 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

2 sin (θ) - 1 = 0 : 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

θ, tan (θ) = 1

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

csc (3 θ) = 6 sin (3 θ), 0 < θ < 2 π

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

(cot (θ) + 1) (csc (θ) - 1) = 0

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