C3 Question

Can someone give me a hand solving this please?

\sin 2\theta = \sin \theta

2\sin \theta \cos \theta = \sin \theta

At this point, why can you not divide through by

\sin \theta

to get:

\cos \theta = \frac{1}{2}

and have the solution as

\cos^{-1} (\frac{1}{2})

Thanks.

Reply 1

TeeEm

Original post by Substitution

Can someone give me a hand solving this please?

\sin 2\theta = \sin \theta

2\sin \theta \cos \theta = \sin \theta

At this point, why can you not divide through by

\sin \theta

to get:

\cos \theta = \frac{1}{2}

and have the solution as

\cos^{-1} (\frac{1}{2})

Thanks.

because sinθ can equal zero, so potentially you are dividing by zero.

In the context of this question you will lose the solutions of the equation sinθ = 0

Reply 2

Substitution

Original post by TeeEm

because sinθ can equal zero, so potentially you are dividing by zero.

In the context of this question you will lose the solutions of the equation sinθ = 0