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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sin(2θ)=cos(θ),2

sin (2 θ) = cos (θ), 2

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) + 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, 2 :

No Solution

2 sin (θ) - 1 = 0, 2 :

No Solution

Combine all the solutions

No Solution for θ \in R

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

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

w, 100 cos (10 w) = 50

cos (θ) = \frac{8}{25}

sin (2 x) = π

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}