Question

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

Answer 1

Answer:

$\theta=0,\pi,\frac{3\pi}{2}$ on the interval 0 ≤ θ <2π

Step-by-step 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:

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

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

$\theta=0,\pi,\frac{3\pi}{2}$