Trigonometry
Trigonometric equation — counting solutions
MJAT_TS5_P1
Grade 12
Question:
Suppose $a$ is a real number such that the equation $a(\sin x+\sin 2x)=\sin 3x$ has more than one solution in the interval $(0,\pi)$. The number of integral values of $a$ satisfying the given condition is:
Step-by-Step Solution
Key Concept: Write $\sin 3x = \sin x(4\cos^2 x-1)$ and $\sin 2x=2\sin x\cos x$. Factor out $\sin x$: for $x\neq 0,\pi$, divide by $\sin x$: $a(1+2\cos x)=4\cos^2 x-1=(2\cos x-1)(2\cos x+1)$. If $2\cos x+1=0$ ($\cos x=-1/2$): LHS$=0$, RHS$=0$ ✓ — always a solution. Otherwise: $a=2\cos x-1$. This has more than one solution in $(0,\pi)$ for certain $a$.
Integral values of $a$: $\mathbf{2}$ values (from analysis). Answer: C.
Correct Answer: C