Question

Solve sin θ +1 = cos 2 θ on the interval 0 ≤ θ <2π

Answer 1

`\theta=0,\pi,(3\pi)/(2)` on the interval 0 ≤ θ <2π

Explanation:

We have
`\sin \theta+1=\cos^2\theta`

We use the Pythagorean identity and substitute
`cos^2\theta=1-sin^2\theta`

`\implies \sin \theta+1=1-\sin^2\theta`

`\implies \sin \theta+1=1-\sin^2\theta`

`\implies \sin^2\theta+\sin \theta=1-1`

`\implies \sin^2\theta+\sin \theta=0`

Factor to get:

`\implies \sin\theta(\sin \theta+1)=0`

`\implies \sin\theta=0\:or\:\sin \theta=-1`

`\theta=0,\pi,(3\pi)/(2)`