Using an equivalent analytic representation, what are all solutions to $\cos(2x)+\sin x=0$ on $0\le x<2\pi$?

Answer and derivation

Correct answer: B

The equivalent form $(1-\sin x)(2\sin x+1)=0$ gives $\sin x=1$ or $\sin x=-1/2$, producing the three solutions in choice B.

1. Use $\cos(2x)=1-2\sin^2x$ to write the equation as $1-2\sin^2x+\sin x=0$.
2. Factor the equivalent expression as $(1-\sin x)(2\sin x+1)=0$.
3. On the stated interval, $\sin x=1$ gives $\pi/2$, while $\sin x=-1/2$ gives $7\pi/6$ and $11\pi/6$.

Why the other options fail

A

This keeps the sin x=1 branch but loses both solutions from sin x=-1/2.

B

Correct solutions from the two factors after rewriting cosine of the double angle.

C

These use sin x=1/2 and sin x=-1, which are not the factored values.

D

At these inputs sin x=0 and cos(2x)=1, so the original left side is 1.

Common misconception: An equivalent identity is useful only if every resulting factor and every domain-valid branch is retained.

China/AP bridge: China-course bridge: algebraic factoring is familiar; AP newly expects students to choose an equivalent trigonometric representation that exposes solutions; unlike applying an inverse key to the mixed original form, rewrite to one trig function first; a common pitfall is discarding a factor or a quadrant.